properties of limits homework 1.3

Calculus 10th Edition

By larson, ron; edwards, bruce h., chapter 1 - limits and their properties - 1.3 exercises - page 68: 69, work step by step, update this answer.

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1.3 Finding Limits Analytically

In Section 1.1 we explored the concept of the limit without a strict definition, meaning we could only make approximations. In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. Thus far, our method of finding a limit is (1) make a really good approximation either graphically or numerically, and (2) verify our approximation is correct using an ϵ - δ proof.

Recognizing that ϵ - δ proofs are cumbersome, this section gives a series of theorems which allow us to find limits much more quickly and intuitively.

Suppose that lim x → 2 ⁡ f ⁢ ( x ) = 2 and lim x → 2 ⁡ g ⁢ ( x ) = 3 . What is lim x → 2 ⁡ ( f ⁢ ( x ) + g ⁢ ( x ) ) ? Intuition tells us that the limit should be 5, as we expect limits to behave in a nice way. The following theorem states that already established limits do behave nicely.

Theorem 1.3.1 Basic Limit Properties

Let b , c , L and K be real numbers, let n be a positive integer, and let f and g be functions with the following limits:

The following limits hold. 1.   Constants: lim x → c ⁡ b = b 2.   Identity: lim x → c ⁡ x = c 3.   Sums/Differences: lim x → c ⁡ ( f ⁢ ( x ) ± g ⁢ ( x ) ) = L ± K 4.   Scalar Multiples: lim x → c ⁡ b ⋅ f ⁢ ( x ) = b ⁢ L 5.   Products: lim x → c ⁡ f ⁢ ( x ) ⋅ g ⁢ ( x ) = L ⁢ K 6.   Quotients: lim x → c ⁡ f ⁢ ( x ) / g ⁢ ( x ) = L / K , ( K ≠ 0 ) 7.   Powers: lim x → c ⁡ [ f ⁢ ( x ) ] n = L n 8.   Roots: lim x → c ⁡ f ⁢ ( x ) n = L n   (when n is odd or L ≥ 0 )

We will now prove the Sum Property using the formal definition of a limit from the previous section. We know that lim x → c ⁡ f ⁢ ( x ) = L and lim x → c ⁡ g ⁢ ( x ) = K . We want to show that lim x → c ⁡ ( f ⁢ ( x ) + g ⁢ ( x ) ) = L + K .

We must show that given any ϵ > 0 , we can find a δ > 0 such that

We know lim x → c ⁡ f ⁢ ( x ) = L . So for any ϵ 1 > 0 , we can find δ 1 > 0 such that if 0 < | x - c | < δ 1 , then | f ⁢ ( x ) - L | < ϵ 1 . Similarly we know lim x → c ⁡ g ⁢ ( x ) = K so for any ϵ 2 > 0 , we can find δ 2 > 0 such that if 0 < | x - c | < δ 2 , then | g ⁢ ( x ) - K | < ϵ 2 . We will let both ϵ 1 and ϵ 2 be ϵ 2 . Now, we have a δ 1 > 0 and a δ 2 > 0 such that:

We will choose δ = min ⁡ ( δ 1 , δ 2 ) > 0 . If 0 < | x - c | < δ , then | f ⁢ ( x ) - L | < ϵ 2 and | g ⁢ ( x ) - K | < ϵ 2 . Add the two inequalities together so that

We will now use the triangle inequality: | A + B | ≤ | A | + | B | .

Thus | ( f ⁢ ( x ) + g ⁢ ( x ) ) - ( L + K ) | < ϵ , which is what we were trying to show. ∎

The other Basic Limit Properties can be proven in a similar way and are left for the reader. Our next theorem requires a few more conditions.

Theorem 1.3.2 Limits of Composition

Suppose that

Then lim x → c ⁡ g ⁢ ( f ⁢ ( x ) ) = K .

We apply the theorem to an example.

Example 1.3.1 Using basic limit properties

Find the following limits:

Using the Sum/Difference rule, we know that lim x → 2 ⁡ ( f ⁢ ( x ) + g ⁢ ( x ) ) = 2 + 3 = 5 .

Using the Scalar Multiple and Sum/Difference rules, we find that lim x → 2 ⁡ ( 5 ⁢ f ⁢ ( x ) + g ⁢ ( x ) 2 ) = 5 ⋅ 2 + 3 2 = 19 .

Here we combine the Power, Scalar Multiple, Sum/Difference and Constant Rules. We show quite a few steps, but in general these can be omitted:

Part 3 of the previous example demonstrates how the limit of a quadratic polynomial can be determined using the properties of Theorem 1.3.1 . Not only that, recognize that

i.e., the limit at 2 was found just by plugging 2 into the function. This holds true for all polynomials, and also for rational functions (which are quotients of polynomials), as stated in the following theorem.

Theorem 1.3.3 Limits of Polynomial and Rational Functions

Let p ⁢ ( x ) and q ⁢ ( x ) be polynomials and c a real number. Then:

lim x → c ⁡ p ⁢ ( x ) = p ⁢ ( c )

lim x → c ⁡ p ⁢ ( x ) q ⁢ ( x ) = p ⁢ ( c ) q ⁢ ( c ) , where q ⁢ ( c ) ≠ 0 .

Example 1.3.2 Finding a limit of a rational function

Using Theorem 1.3.3 , find

Solution Using Theorem 1.3.3 , we can quickly state that

It was likely frustrating in Section 1.2 to do a lot of work to prove that

as it seemed fairly obvious. The previous theorems state that many functions behave in such an “obvious” fashion, as demonstrated by the rational function in Example 1.3.2 .

Polynomial and rational functions are not the only functions to behave in such a predictable way. The following theorem gives a list of functions whose behavior is particularly “nice” in terms of limits. In the next section, we will give a formal name to these functions that behave “nicely.”

Theorem 1.3.4 Limits of Basic Functions

Let c be a real number in the domain of the given function and let n be a positive integer. The following limits hold:

lim x → c ⁡ sin ⁡ x = sin ⁡ c

lim x → c ⁡ cos ⁡ x = cos ⁡ c

lim x → c ⁡ tan ⁡ x = tan ⁡ c

lim x → c ⁡ csc ⁡ x = csc ⁡ c

lim x → c ⁡ sec ⁡ x = sec ⁡ c

lim x → c ⁡ cot ⁡ x = cot ⁡ c

lim x → c ⁡ a x = a c ( a > 0 )

lim x → c ⁡ ln ⁡ x = ln ⁡ c

lim x → c ⁡ x n = c n

Many times, we will combine this theorem with Theorems 1.3.1 and 1.3.2 . If our expression can be built up from the pieces in those theorems, then we can quickly evaluate the limit.

Example 1.3.3 Evaluating limits analytically

Evaluate the following limits.

lim x → π ⁡ cos ⁡ x

lim x → 3 ⁡ ( sec 2 ⁡ x - tan 2 ⁡ x )

lim x → π 2 ⁡ cos ⁡ x ⁢ sin ⁡ x

lim x → 1 ⁡ e ln ⁡ x

lim x → 0 ⁡ sin ⁡ x x

This is a straightforward application of Theorem 1.3.4 : lim x → π ⁡ cos ⁡ x = cos ⁡ π = - 1 .

We can approach this in at least two ways. First, by directly applying Theorems 1.3.1 and 1.3.4 , we have:

Using the Pythagorean Theorem, this last expression is 1; therefore

We can also use the Pythagorean Theorem from the start:

using the Constant limit rule. Either way, we find the limit is 1.

Applying the Product limit rule of Theorem 1.3.1 and Theorem 1.3.4 gives

Again, we can approach this in two ways. First, we can use the exponential/logarithmic identity that e ln ⁡ x = x and evaluate lim x → 1 ⁡ e ln ⁡ x = lim x → 1 ⁡ x = 1 .

We can also use Theorem 1.3.2 . Using Theorem 1.3.4 , we have lim x → 1 ⁡ ln ⁡ x = ln ⁡ 1 = 0 . Applying the Composition rule,

Both approaches are valid, giving the same result.

We encountered this limit in Section 1.1 . Applying our theorems, we attempt to find the limit as

This, of course, violates a condition of Theorem 1.3.1 , as the limit of the denominator is not allowed to be 0. Therefore, we are still unable to evaluate this limit with tools we currently have at hand.

The section could have been titled “Using Known Limits to Find Unknown Limits.” By knowing certain limits of functions, we can find limits involving sums, products, powers, etc., of these functions. We further the development of such comparative tools with the Squeeze Theorem, a clever and intuitive way to find the value of some limits.

Before stating this theorem formally, suppose we have functions f , g and h where g always takes on values between f and h ; that is, for all x in an interval,

If f and h have the same limit at c , and g is always “squeezed” between them, then g must have the same limit as well. That is what the Squeeze Theorem states, as illustrated in Figure 1.3.1 .

Theorem 1.3.5 Squeeze Theorem

Let f , g and h be functions on open intervals I and J on either side of c such that for all x in I and J ,

It can take some work to figure out appropriate functions with which to “squeeze” the given function of which you are trying to evaluate a limit. However, that is generally the only place work is necessary; the theorem makes the “evaluating the limit part” very simple.

We use the Squeeze Theorem in the following example to finally prove that lim x → 0 ⁡ sin ⁡ x x = 1 .

Example 1.3.4 Using the Squeeze Theorem

Use the Squeeze Theorem to show that

Solution We will begin by considering the unit circle. Each point on the unit circle has coordinates ( cos ⁡ θ , sin ⁡ θ ) for some angle θ as shown in Figure 1.3.2 . Using similar triangles, we can extend the line from the origin through the point to the point ( 1 , tan ⁡ θ ) , as shown. (Here we are assuming that 0 ≤ θ ≤ π / 2 . Later we will show that we can also consider θ ≤ 0 .)

Figure 1.3.2 shows three regions have been constructed in the first quadrant, two triangles and a sector of a circle, which are also drawn below. The area of the large triangle is 1 2 ⁢ tan ⁡ θ ; the area of the sector is θ / 2 ; the area of the triangle contained inside the sector is 1 2 ⁢ sin ⁡ θ . It is then clear from the diagram that

Multiply all terms by 2 sin ⁡ θ , giving

Taking reciprocals reverses the inequalities, giving

(These inequalities hold for all values of θ near 0, even negative values, since cos ⁡ ( - θ ) = cos ⁡ θ and sin ⁡ ( - θ ) = - sin ⁡ θ .)

Now take limits.

Clearly this means that lim θ → 0 ⁡ sin ⁡ θ θ = 1 .

Two notes about the previous example are worth mentioning. First, one might be discouraged by this application, thinking “I would never have come up with that on my own. This is too hard!” Don’t be discouraged; within this text we will guide you in your use of the Squeeze Theorem. As one gains mathematical maturity, clever proofs like this are easier and easier to create.

Second, this limit tells us more than just that as x approaches 0, sin ⁡ ( x ) / x approaches 1. Both x and sin ⁡ x are approaching 0, but the ratio of x and sin ⁡ x approaches 1, meaning that they are approaching 0 in essentially the same way. Another way of viewing this is: for small x , the functions y = x and y = sin ⁡ x are essentially indistinguishable.

We include this special limit, along with three others, in the following theorem.

Theorem 1.3.6 Special Limits

lim x → 0 ⁡ sin ⁡ x x = 1

lim x → 0 ⁡ cos ⁡ x - 1 x = 0

lim x → 0 ⁡ ( 1 + x ) 1 x = e

lim x → 0 ⁡ e x - 1 x = 1

A short word on how to interpret the latter three limits. We know that as x goes to 0, cos ⁡ x goes to 1. So, in the second limit, both the numerator and denominator are approaching 0. However, since the limit is 0, we can interpret this as saying that “ cos ⁡ x is approaching 1 faster than x is approaching 0.”

In the third limit, inside the parentheses we have an expression that is approaching 1 (though never equaling 1), and we know that 1 raised to any power is still 1. At the same time, the power is growing toward infinity. What happens to a number near 1 raised to a very large power? In this particular case, the result approaches Euler’s number, e , approximately 2.718 .

In the fourth limit, we see that as x → 0 , e x approaches 1 “just as fast” as x → 0 , resulting in a limit of 1.

Our final theorem for this section will be motivated by the following example.

Example 1.3.5 Using algebra to evaluate a limit

Evaluate the following limit:

Solution We would like to apply Theorems 1.3.1 and 1.3.3 and substitute 1 for x in the quotient. This gives:

an indeterminate form. We cannot apply the Theorem 1.3.1 because the denominator is 0.

By graphing the function, as in Figure 1.3.3 , we see that the function seems to be linear, implying that the limit should be easy to evaluate. Recognize that the numerator of our quotient can be factored:

The function is not defined when x = 1 , but for all other x ,

Clearly lim x → 1 ⁡ ( x + 1 ) = 2 . Recall that when considering limits, we are not concerned with the value of the function at 1, only the value that the function approaches as x approaches 1. Since ( x 2 - 1 ) / ( x - 1 ) and x + 1 are the same at all points except x = 1 , they both approach the same value as x approaches 1. Therefore we can conclude that

The key to the above example is that the functions y = ( x 2 - 1 ) / ( x - 1 ) and y = x + 1 are identical except at x = 1 . Since limits describe a value the function is approaching, not the value the function actually attains, the limits of the two functions are always equal.

Theorem 1.3.7 Limits of Functions Equal At All But One Point

Let g ⁢ ( x ) = f ⁢ ( x ) for all x in an open interval, except possibly at c , and let lim x → c ⁡ g ⁢ ( x ) = L for some real number L . Then

The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form g ⁢ ( x ) / f ⁢ ( x ) and directly evaluating the limit lim x → c ⁡ g ⁢ ( x ) f ⁢ ( x ) returns “0/0”, then ( x - c ) is a factor of both g ⁢ ( x ) and f ⁢ ( x ) . One can then use algebra to factor this term out, divide, then apply Theorem 1.3.7 . Some useful algebraic techniques to rewrite functions that return an indeterminate form when evaluating a limit are:

factoring and dividing out common factors,

rationalizing the numerator or denominator,

simplifying the expression, and

finding a common denominator.

We will demonstrate some of these techniques in the following examples.

Example 1.3.6 Evaluating a limit using Theorem 1.3.7

Evaluate lim x → 3 ⁡ x 3 - 2 ⁢ x 2 - 5 ⁢ x + 6 2 ⁢ x 3 + 3 ⁢ x 2 - 32 ⁢ x + 15 .

Solution We begin by attempting to apply Theorems 1.3.1 and 1.3.4 and substituting 3 for x . This returns the familiar indeterminate form of “0/0”. Since the numerator and denominator are each polynomials, we know that ( x - 3 ) is factor of each. Using whatever method is most comfortable to you, factor out ( x - 3 ) from each (using polynomial division, synthetic division, a computer algebra system, etc.). We find that

We can divide the ( x - 3 ) terms as long as x ≠ 3 . Using Theorem 1.3.7 we conclude:

Example 1.3.7 Evaluating a limit by rationalizing

Evaluate lim x → 0 ⁡ x + 4 - 2 x .

Solution We begin by applying Theorem 1.3.4 and substituting 2 for x . This returns the familiar indeterminate form of “0/0”. We see the radical in the numerator so we will rationalize the numerator. Using Theorem 1.3.7 we find that

Notice that we didn’t distribute the denominator in the second line. Generally speaking, when we are hoping to divide out a factor in a fraction we will need to undo any distributing that we may have prematurely done.

We end this section by revisiting a limit first seen in Section 1.1 , a limit of a difference quotient. Let f ⁢ ( x ) = - 1.5 ⁢ x 2 + 11.5 ⁢ x ; we approximated the limit lim h → 0 ⁡ f ⁢ ( 1 + h ) - f ⁢ ( 1 ) h ≈ 8.5 . We formally evaluate this limit in the following example.

Example 1.3.8 Evaluating the limit of a difference quotient

Let f ⁢ ( x ) = - 1.5 ⁢ x 2 + 11.5 ⁢ x ; find lim h → 0 ⁡ f ⁢ ( 1 + h ) - f ⁢ ( 1 ) h .

Solution Since f is a polynomial, our first attempt should be to employ Theorem 1.3.4 and substitute 0 for h . However, we see that this gives us “ 0 / 0 .” Knowing that we have a rational function hints that some algebra will help. Consider the following steps:

This matches our previous approximation.

This section contains several valuable tools for evaluating limits. One of the main results of this section is Theorem 1.3.4 ; it states that many functions that we use regularly behave in a very nice, predictable way. In Section 1.6 we give a name to this nice behavior; we label such functions as continuous. Defining that term will require us to look again at what a limit is and what causes limits to not exist.

Exercises 1.3

Terms and concepts.

Explain in your own words, without using ϵ - δ formality, why lim x → c ⁡ b = b .

Explain in your own words, without using ϵ - δ formality, why lim x → c ⁡ x = c .

What does the text mean when it says that certain functions’ “behavior is ‘nice’ in terms of limits”? What, in particular, is “nice”?

Sketch a graph that visually demonstrates the Squeeze Theorem.

• 5. You are given the following information: (a) lim x → 1 ⁡ f ⁢ ( x ) = 0 (b) lim x → 1 ⁡ g ⁢ ( x ) = 0 (c) lim x → 1 ⁡ f ⁢ ( x ) / g ⁢ ( x ) = 2 What can be said about the relative sizes of f ⁢ ( x ) and g ⁢ ( x ) as x approaches 1?

T/F: lim x → 1 ⁡ ln ⁡ x = 0 . Use a theorem to defend your answer.

Use the following limits to evaluate the limits given in Exercises 7– 14. , where possible. If it is not possible, state so.

lim x → 9 ⁡ ( f ⁢ ( x ) + g ⁢ ( x ) )

lim x → 9 ⁡ ( 3 ⁢ f ⁢ ( x ) / g ⁢ ( x ) )

lim x → 9 ⁡ ( f ⁢ ( x ) - 2 ⁢ g ⁢ ( x ) g ⁢ ( x ) )

lim x → 6 ⁡ ( f ⁢ ( x ) 3 - g ⁢ ( x ) )

lim x → 9 ⁡ g ⁢ ( f ⁢ ( x ) )

lim x → 6 ⁡ f ⁢ ( g ⁢ ( x ) )

lim x → 6 ⁡ g ⁢ ( f ⁢ ( f ⁢ ( x ) ) )

lim x → 6 ⁡ ( f ⁢ ( x ) ⁢ g ⁢ ( x ) - f 2 ⁢ ( x ) + g 2 ⁢ ( x ) )

Use the following limits to evaluate the limits given in Exercises 15– 18. , where possible. If it is not possible, state so.

lim x → 1 ⁡ f ⁢ ( x ) g ⁢ ( x )

lim x → 10 ⁡ cos ⁡ ( g ⁢ ( x ) )

lim x → 1 ⁡ f ⁢ ( x ) ⁢ g ⁢ ( x )

lim x → 1 ⁡ g ⁢ ( 5 ⁢ f ⁢ ( x ) )

In Exercises 19– 40. , evaluate the given limit.

lim x → 3 ⁡ ( x 2 - 3 ⁢ x + 7 )

lim x → π ⁡ ( x - 3 x - 5 ) 7

lim x → π / 4 ⁡ cos ⁡ x ⁢ sin ⁡ x

lim x → 1 ⁡ 2 ⁢ x - 2 x + 4

lim x → 0 ⁡ ln ⁡ x

lim x → 3 ⁡ 4 x 3 - 8 ⁢ x

lim x → π / 6 ⁡ csc ⁡ x

lim x → 0 ⁡ ln ⁡ ( 1 + x )

lim x → π ⁡ x 2 + 3 ⁢ x + 5 5 ⁢ x 2 - 2 ⁢ x - 3

lim x → π ⁡ 3 ⁢ x + 1 1 - x

lim x → 6 ⁡ x 2 - 4 ⁢ x - 12 x 2 - 13 ⁢ x + 42

lim x → 0 ⁡ x 2 + 2 ⁢ x x 2 - 2 ⁢ x

lim x → 2 ⁡ x 2 + 6 ⁢ x - 16 x 2 - 3 ⁢ x + 2

lim x → 2 ⁡ x 2 - 10 ⁢ x + 16 x 2 - x - 2

lim x → - 2 ⁡ x 2 - 5 ⁢ x - 14 x 2 + 10 ⁢ x + 16

lim x → - 1 ⁡ x 2 + 9 ⁢ x + 8 x 2 - 6 ⁢ x - 7

lim t → 9 ⁡ t - 3 t - 9

lim x → 0 ⁡ x 2 + 4 - 2 x 2

lim t → 3 ⁡ 1 t - 1 3 t - 3

lim x → 0 ⁡ 1 x - 1 x 2 + x

lim t → 0 ⁡ ( t - 4 ) 2 - 16 t

lim x → 13 ⁡ x + 3 - 4 x - 13

Use the Squeeze Theorem in Exercises 41– 46. , where appropriate, to evaluate the given limit. Hint: - 1 ≤ sin ⁡ x ≤ 1 and - 1 ≤ cos ⁡ x ≤ 1 .

lim x → 0 ⁡ x ⁢ sin ⁡ ( 1 x )

lim x → 0 ⁡ sin ⁡ x ⁢ cos ⁡ ( 1 x 2 )

lim x → 1 ⁡ f ⁢ ( x ) , where 3 ⁢ x - 2 ≤ f ⁢ ( x ) ≤ x 3 .

lim x → 3 ⁡ f ⁢ ( x ) , where 6 ⁢ x - 9 ≤ f ⁢ ( x ) ≤ x 2 .

lim x → 0 ⁡ x 2 ⁢ cos ⁡ ( 1 x )

lim x → 0 ⁡ x ⁢ ⌊ 1 x ⌋ , where ⌊ y ⌋ is the greatest integer less than or equal to y .

Exercises 47– 52. challenge your understanding of limits that can be evaluated using the knowledge gained in this section.

lim x → 0 ⁡ sin ⁡ 3 ⁢ x x

lim x → 0 ⁡ sin ⁡ 5 ⁢ x 8 ⁢ x

lim x → 0 ⁡ ln ⁡ ( 1 + x ) x

lim x → 0 ⁡ sin ⁡ x x , where x is measured in degrees, not radians.

lim x → 0 ⁡ tan ⁡ 4 ⁢ x tan ⁡ 3 ⁢ x

lim x → 0 ⁡ tan ⁡ 5 ⁢ x sin ⁡ 7 ⁢ x

• 53. Let f ⁢ ( x ) = 0 and g ⁢ ( x ) = x x . (a) Show why lim x → 2 ⁡ f ⁢ ( x ) = 0 . (b) Show why lim x → 0 ⁡ g ⁢ ( x ) = 1 . (c) Show why lim x → 2 ⁡ g ⁢ ( f ⁢ ( x ) ) does not exist. (d) Show why the answer to part (c) does not violate Theorem 1.3.2 .

Verify lim x → 0 ⁡ cos ⁡ x - 1 x = 0  Hint: Multiply by cos ⁡ x + 1 cos ⁡ x + 1

12.2 Finding Limits: Properties of Limits

Learning objectives.

In this section, you will:

• Find the limit of a sum, a difference, and a product.
• Find the limit of a polynomial.
• Find the limit of a power or a root.
• Find the limit of a quotient.

Consider the rational function

The function can be factored as follows:

Does this mean the function f f is the same as the function g ( x ) = x + 1 ? g ( x ) = x + 1 ?

The answer is no. Function f f does not have x = 7 x = 7 in its domain, but g g does. Graphically, we observe there is a hole in the graph of f ( x ) f ( x ) at x = 7 , x = 7 , as shown in Figure 1 and no such hole in the graph of g ( x ) , g ( x ) , as shown in Figure 2 .

So, do these two different functions also have different limits as x x approaches 7?

Not necessarily. Remember, in determining a limit of a function as x x approaches a , a , what matters is whether the output approaches a real number as we get close to x = a . x = a . The existence of a limit does not depend on what happens when x x equals a . a .

Look again at Figure 1 and Figure 2 . Notice that in both graphs, as x x approaches 7, the output values approach 8. This means

Remember that when determining a limit, the concern is what occurs near x = a , x = a , not at x = a . x = a . In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition.

Finding the Limit of a Sum, a Difference, and a Product

Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits , which is a collection of theorems for finding limits.

Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.

Properties of Limits

Let a , k , A , a , k , A , and B B represent real numbers, and f f and g g be functions, such that lim x → a f ( x ) = A lim x → a f ( x ) = A and lim x → a g ( x ) = B . lim x → a g ( x ) = B . For limits that exist and are finite, the properties of limits are summarized in Table 1

Evaluating the Limit of a Function Algebraically

Evaluate lim x → 3 ( 2 x + 5 ) . lim x → 3 ( 2 x + 5 ) .

Evaluate the following limit: lim x → − 12 ( − 2 x + 2 ) . lim x → − 12 ( − 2 x + 2 ) .

Finding the Limit of a Polynomial

Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as x x approaches a a is equivalent to simply evaluating the function for a a .

Given a function containing a polynomial, find its limit.

• Use the properties of limits to break up the polynomial into individual terms.
• Find the limits of the individual terms.
• Add the limits together.
• Alternatively, evaluate the function for a a .

Evaluate lim x → 3 ( 5 x 2 ) . lim x → 3 ( 5 x 2 ) .

Evaluate lim x → 4 ( x 3 − 5 ) . lim x → 4 ( x 3 − 5 ) .

Evaluating the Limit of a Polynomial Algebraically

Evaluate lim x → 5 ( 2 x 3 − 3 x + 1 ) . lim x → 5 ( 2 x 3 − 3 x + 1 ) .

Evaluate the following limit: lim x → − 1 ( x 4 − 4 x 3 + 5 ) . lim x → − 1 ( x 4 − 4 x 3 + 5 ) .

Finding the Limit of a Power or a Root

When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.

Evaluating a Limit of a Power

Evaluate lim x → 2 ( 3 x + 1 ) 5 . lim x → 2 ( 3 x + 1 ) 5 .

We will take the limit of the function as x x approaches 2 and raise the result to the 5 th power.

Evaluate the following limit: lim x → − 4 ( 10 x + 36 ) 3 . lim x → − 4 ( 10 x + 36 ) 3 .

If we can’t directly apply the properties of a limit, for example in lim x → 2 ( x 2 + 6 x + 8 x − 2 ) lim x → 2 ( x 2 + 6 x + 8 x − 2 ) , can we still determine the limit of the function as x x approaches a a ?

Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function.

Finding the Limit of a Quotient

Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.

Given the limit of a function in quotient form, use factoring to evaluate it.

• Factor the numerator and denominator completely.
• Simplify by dividing any factors common to the numerator and denominator.
• Evaluate the resulting limit, remembering to use the correct domain.

Evaluating the Limit of a Quotient by Factoring

Evaluate lim x → 2 ( x 2 − 6 x + 8 x − 2 ) . lim x → 2 ( x 2 − 6 x + 8 x − 2 ) .

Factor where possible, and simplify.

When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated.

Notice, the function

is equivalent to the function

Notice that the limit exists even though the function is not defined at x  = 2 . x  = 2 .

Evaluate the following limit: lim x → 7 ( x 2 − 11 x + 28 7 − x ) . lim x → 7 ( x 2 − 11 x + 28 7 − x ) .

Evaluating the Limit of a Quotient by Finding the LCD

Evaluate lim x → 5 ( 1 x − 1 5 x − 5 ) . lim x → 5 ( 1 x − 1 5 x − 5 ) .

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

When determining the limit of a rational function that has terms added or subtracted in either the numerator or denominator, the first step is to find the common denominator of the added or subtracted terms; then, convert both terms to have that denominator, or simplify the rational function by multiplying numerator and denominator by the least common denominator. Then check to see if the resulting numerator and denominator have any common factors.

Evaluate lim x → − 5 ( 1 5 + 1 x 10 + 2 x ) . lim x → − 5 ( 1 5 + 1 x 10 + 2 x ) .

Given a limit of a function containing a root, use a conjugate to evaluate.

• If the quotient as given is not in indeterminate ( 0 0 ) ( 0 0 ) form, evaluate directly.
• Otherwise, rewrite the sum (or difference) of two quotients as a single quotient, using the least common denominator (LCD) .
• If the numerator includes a root, rationalize the numerator; multiply the numerator and denominator by the conjugate of the numerator. Recall that a ± b a ± b are conjugates.
• Evaluate the resulting limit.

Evaluating a Limit Containing a Root Using a Conjugate

Evaluate lim x → 0 ( 25 − x − 5 x ) . lim x → 0 ( 25 − x − 5 x ) .

lim x → 0 ( 25 − x − 5 x ) = lim x → 0 ( ( 25 − x − 5 ) x ⋅ ( 25 − x + 5 ) ( 25 − x + 5 ) ) Multiply numerator and denominator by the conjugate .                                 = lim x → 0 ( ( 25 − x ) − 25 x ( 25 − x + 5 ) ) Multiply:  ( 25 − x − 5 ) ⋅ ( 25 − x + 5 ) = ( 25 − x ) − 25.                                 = lim x → 0 ( − x x ( 25 − x + 5 ) ) Combine like terms .                                 = lim x → 0 ( − x x ( 25 − x + 5 ) ) Simplify  − x x = − 1.                                 = − 1 25 − 0 + 5 Evaluate .                                 = − 1 5 + 5 = − 1 10 lim x → 0 ( 25 − x − 5 x ) = lim x → 0 ( ( 25 − x − 5 ) x ⋅ ( 25 − x + 5 ) ( 25 − x + 5 ) ) Multiply numerator and denominator by the conjugate .                                 = lim x → 0 ( ( 25 − x ) − 25 x ( 25 − x + 5 ) ) Multiply:  ( 25 − x − 5 ) ⋅ ( 25 − x + 5 ) = ( 25 − x ) − 25.                                 = lim x → 0 ( − x x ( 25 − x + 5 ) ) Combine like terms .                                 = lim x → 0 ( − x x ( 25 − x + 5 ) ) Simplify  − x x = − 1.                                 = − 1 25 − 0 + 5 Evaluate .                                 = − 1 5 + 5 = − 1 10

When determining a limit of a function with a root as one of two terms where we cannot evaluate directly, think about multiplying the numerator and denominator by the conjugate of the terms.

Evaluate the following limit: lim h → 0 ( 16 − h − 4 h ) . lim h → 0 ( 16 − h − 4 h ) .

Evaluating the Limit of a Quotient of a Function by Factoring

Evaluate lim x → 4 ( 4 − x x − 2 ) . lim x → 4 ( 4 − x x − 2 ) .

lim x → 4 ( 4 − x x − 2 ) = lim x → 4 ( ( 2 + x ) ( 2 − x ) x − 2 ) Factor.                        = lim x → 4 ( ( 2 + x ) ( 2 − x ) − ( 2 − x ) ) Factor  −1  out of the denominator . Simplify .                        = lim x → 4 − ( 2 + x ) Evaluate .                        = − ( 2 + 4 )                        = − 4 lim x → 4 ( 4 − x x − 2 ) = lim x → 4 ( ( 2 + x ) ( 2 − x ) x − 2 ) Factor.                        = lim x → 4 ( ( 2 + x ) ( 2 − x ) − ( 2 − x ) ) Factor  −1  out of the denominator . Simplify .                        = lim x → 4 − ( 2 + x ) Evaluate .                        = − ( 2 + 4 )                        = − 4

Multiplying by a conjugate would expand the numerator; look instead for factors in the numerator. Four is a perfect square so that the numerator is in the form

and may be factored as

Evaluate the following limit: lim x → 3 ( x − 3 x − 3 ) . lim x → 3 ( x − 3 x − 3 ) .

Given a quotient with absolute values, evaluate its limit.

• Try factoring or finding the LCD.
• If the limit cannot be found, choose several values close to and on either side of the input where the function is undefined.
• Use the numeric evidence to estimate the limits on both sides.

Evaluating the Limit of a Quotient with Absolute Values

Evaluate lim x → 7 | x − 7 | x − 7 . lim x → 7 | x − 7 | x − 7 .

The function is undefined at x = 7 , x = 7 , so we will try values close to 7 from the left and the right.

Left-hand limit: | 6.9 − 7 | 6.9 − 7 = | 6.99 − 7 | 6.99 − 7 = | 6.999 − 7 | 6.999 − 7 = − 1 | 6.9 − 7 | 6.9 − 7 = | 6.99 − 7 | 6.99 − 7 = | 6.999 − 7 | 6.999 − 7 = − 1

Right-hand limit: | 7.1 − 7 | 7.1 − 7 = | 7.01 − 7 | 7.01 − 7 = | 7.001 − 7 | 7.001 − 7 = 1 | 7.1 − 7 | 7.1 − 7 = | 7.01 − 7 | 7.01 − 7 = | 7.001 − 7 | 7.001 − 7 = 1

Since the left- and right-hand limits are not equal, there is no limit.

Evaluate lim x → 6 + 6 − x | x − 6 | . lim x → 6 + 6 − x | x − 6 | .

Access the following online resource for additional instruction and practice with properties of limits.

• Determine a Limit Analytically

12.2 Section Exercises

Give an example of a type of function f f whose limit, as x x approaches a , a , is f ( a ) . f ( a ) .

When direct substitution is used to evaluate the limit of a rational function as x x approaches a a and the result is f ( a ) = 0 0 , f ( a ) = 0 0 , does this mean that the limit of f f does not exist?

What does it mean to say the limit of f ( x ) , f ( x ) , as x x approaches c , c , is undefined?

For the following exercises, evaluate the limits algebraically.

lim x → 0 ( 3 ) lim x → 0 ( 3 )

lim x → 2 ( − 5 x x 2 − 1 ) lim x → 2 ( − 5 x x 2 − 1 )

lim x → 2 ( x 2 − 5 x + 6 x + 2 ) lim x → 2 ( x 2 − 5 x + 6 x + 2 )

lim x → 3 ( x 2 − 9 x − 3 ) lim x → 3 ( x 2 − 9 x − 3 )

lim x → − 1 ( x 2 − 2 x − 3 x + 1 ) lim x → − 1 ( x 2 − 2 x − 3 x + 1 )

lim x → 3 2 ( 6 x 2 − 17 x + 12 2 x − 3 ) lim x → 3 2 ( 6 x 2 − 17 x + 12 2 x − 3 )

lim x → − 7 2 ( 8 x 2 + 18 x − 35 2 x + 7 ) lim x → − 7 2 ( 8 x 2 + 18 x − 35 2 x + 7 )

lim x → 3 ( x 2 − 9 x − 5 x + 6 ) lim x → 3 ( x 2 − 9 x − 5 x + 6 )

lim x → − 3 ( − 7 x 4 − 21 x 3 − 12 x 4 + 108 x 2 ) lim x → − 3 ( − 7 x 4 − 21 x 3 − 12 x 4 + 108 x 2 )

lim x → 3 ( x 2 + 2 x − 3 x − 3 ) lim x → 3 ( x 2 + 2 x − 3 x − 3 )

lim h → 0 ( ( 3 + h ) 3 − 27 h ) lim h → 0 ( ( 3 + h ) 3 − 27 h )

lim h → 0 ( ( 2 − h ) 3 − 8 h ) lim h → 0 ( ( 2 − h ) 3 − 8 h )

lim h → 0 ( ( h + 3 ) 2 − 9 h ) lim h → 0 ( ( h + 3 ) 2 − 9 h )

lim h → 0 ( 5 − h − 5 h ) lim h → 0 ( 5 − h − 5 h )

lim x → 0 ( 3 − x − 3 x ) lim x → 0 ( 3 − x − 3 x )

lim x → 9 ( x 2 − 81 3 − x ) lim x → 9 ( x 2 − 81 3 − x )

lim x → 1 ( x − x 2 1 − x ) lim x → 1 ( x − x 2 1 − x )

lim x → 0 ( x 1 + 2 x − 1 ) lim x → 0 ( x 1 + 2 x − 1 )

lim x → 1 2 ( x 2 − 1 4 2 x − 1 ) lim x → 1 2 ( x 2 − 1 4 2 x − 1 )

lim x → 4 ( x 3 − 64 x 2 − 16 ) lim x → 4 ( x 3 − 64 x 2 − 16 )

lim x → 2 − ( | x − 2 | x − 2 ) lim x → 2 − ( | x − 2 | x − 2 )

lim x → 2 + ( | x − 2 | x − 2 ) lim x → 2 + ( | x − 2 | x − 2 )

lim x → 2 ( | x − 2 | x − 2 ) lim x → 2 ( | x − 2 | x − 2 )

lim x → 4 − ( | x − 4 | 4 − x ) lim x → 4 − ( | x − 4 | 4 − x )

lim x → 4 + ( | x − 4 | 4 − x ) lim x → 4 + ( | x − 4 | 4 − x )

lim x → 4 ( | x − 4 | 4 − x ) lim x → 4 ( | x − 4 | 4 − x )

lim x → 2 ( − 8 + 6 x − x 2 x − 2 ) lim x → 2 ( − 8 + 6 x − x 2 x − 2 )

For the following exercise, use the given information to evaluate the limits: lim x → c f ( x ) = 3 , lim x → c f ( x ) = 3 , lim x → c g ( x ) = 5 lim x → c g ( x ) = 5 .

lim x → c [ 2 f ( x ) + g ( x ) ] lim x → c [ 2 f ( x ) + g ( x ) ]

lim x → c [ 3 f ( x ) + g ( x ) ] lim x → c [ 3 f ( x ) + g ( x ) ]

lim x → c f ( x ) g ( x ) lim x → c f ( x ) g ( x )

For the following exercises, evaluate the following limits.

lim x → 2 cos ( π x ) lim x → 2 cos ( π x )

lim x → 2 sin ( π x ) lim x → 2 sin ( π x )

f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 + f ( x ) f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 + f ( x )

f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 − f ( x ) f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 − f ( x )

f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 f ( x ) f ( x ) = { 2 x 2 + 2 x + 1 , x ≤ 0 x − 3 , x > 0 ;  lim x → 0 f ( x )

lim x → 4 x + 5 − 3 x − 4 lim x → 4 x + 5 − 3 x − 4

lim x → 2 + ( 2 x − [ x ] ) lim x → 2 + ( 2 x − [ x ] )

lim x → 2 x + 7 − 3 x 2 − x − 2 lim x → 2 x + 7 − 3 x 2 − x − 2

lim x → 3 + x 2 x 2 − 9 lim x → 3 + x 2 x 2 − 9

For the following exercises, find the average rate of change f ( x + h ) − f ( x ) h . f ( x + h ) − f ( x ) h .

f ( x ) = x + 1 f ( x ) = x + 1

f ( x ) = 2 x 2 − 1 f ( x ) = 2 x 2 − 1

f ( x ) = x 2 + 3 x + 4 f ( x ) = x 2 + 3 x + 4

f ( x ) = x 2 + 4 x − 100 f ( x ) = x 2 + 4 x − 100

f ( x ) = 3 x 2 + 1 f ( x ) = 3 x 2 + 1

f ( x ) = cos ( x ) f ( x ) = cos ( x )

f ( x ) = 2 x 3 − 4 x f ( x ) = 2 x 3 − 4 x

f ( x ) = 1 x f ( x ) = 1 x

f ( x ) = 1 x 2 f ( x ) = 1 x 2

f ( x ) = x f ( x ) = x

Find an equation that could be represented by Figure 3 .

Find an equation that could be represented by Figure 4 .

For the following exercises, refer to Figure 5 .

What is the right-hand limit of the function as x x approaches 0?

What is the left-hand limit of the function as x x approaches 0?

Real-World Applications

The position function s ( t ) = − 16 t 2 + 144 t s ( t ) = − 16 t 2 + 144 t gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval [ 1 , 2 ] [ 1 , 2 ] .

The height of a projectile is given by s ( t ) = − 64 t 2 + 192 t s ( t ) = − 64 t 2 + 192 t Find the average rate of change of the height from t = 1 t = 1 second to t = 1.5 t = 1.5 seconds.

The amount of money in an account after t t years compounded continuously at 4.25% interest is given by the formula A = A 0 e 0.0425 t , A = A 0 e 0.0425 t , where A 0 A 0 is the initial amount invested. Find the average rate of change of the balance of the account from t = 1 t = 1 year to t = 2 t = 2 years if the initial amount invested is $1,000.00.

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Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/12-2-finding-limits-properties-of-limits

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1.3: The Limit of a Function

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• Page ID 89706

• Joel Feldman, Andrew Rechnitzer and Elyse Yeager
• University of British Columbia

Before we come to definitions, let us start with a little notation for limits.

Definition 1.3.1.

We will often write

\begin{gather*} \lim_{x \to a} f(x) = L \end{gather*}

which should be read as

The limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\text{.}\)

The notation is just shorthand — we don't want to have to write out long sentences as we do our mathematics. Whenever you see these symbols you should think of that sentence.

This shorthand also has the benefit of being mathematically precise (we'll see this later), and (almost) independent of the language in which the author is writing. A mathematician who does not speak English can read the above formula and understand exactly what it means.

In mathematics, like most languages, there is usually more than one way of writing things and we can also write the above limit as

\begin{gather*} f(x) \to L \mbox{ as } x \to a \end{gather*}

This can also be read as above, but also as

\(f(x)\) goes to \(L\) as \(x\) goes to \(a\)

They mean exactly the same thing in mathematics, even though they might be written, read and said a little differently.

To arrive at the definition of limit, we want to start 1 with a very simple example.

Example 1.3.2. A simple limit.

Consider the following function.

\begin{align*} f(x) &= \begin{cases} 2x & x \lt 3\\ 9 & x=3\\ 2x & x \gt 3 \end{cases} \end{align*}

This is an example of a piece-wise function 2  . That is, a function defined in several pieces, rather than as a single formula. We evaluate the function at a particular value of \(x\) on a case-by-case basis. Here is a sketch of it

Notice the two circles in the plot. One is open, \(\circ\) and the other is closed \(\bullet\text{.}\)

• A filled circle has quite a precise meaning — a filled circle at \((x,y)\) means that the function takes the value \(f(x) = y\text{.}\)
• An open circle is a little harder — an open circle at \((3,6)\) means that the point \((3,6)\) is not on the graph of \(y=f(x)\text{,}\) i.e. \(f(3) \neq 6\text{.}\) We should only use the open circle where it is absolutely necessary in order to avoid confusion.

This function is quite contrived, but it is a very good example to start working with limits more systematically. Consider what the function does close to \(x=3\text{.}\) We already know what happens exactly at \(3\) — \(f(x)=9\) — but I want to look at how the function behaves very close to \(x=3\text{.}\) That is, what does the function do as we look at a point \(x\) that gets closer and closer to \(x=3\text{.}\)

If we plug in some numbers very close to \(3\) (but not exactly 3) into the function we see the following:

So as \(x\) moves closer and closer to 3, without being exactly 3, we see that the function moves closer and closer to 6. We can write this as

\begin{align*} \lim_{x \to 3} f(x) &= 6 \end{align*}

The limit as \(x\) approaches \(3\) of \(f(x)\) is \(6\text{.}\)

So for \(x\) very close to \(3\text{,}\) without being exactly 3, the function is very close to \(6\) — which is a long way from the value of the function exactly at \(3\text{,}\) \(f(3)=9\text{.}\) Note well that the behaviour of the function as \(x\) gets very close to 3 does not depend on the value of the function at 3.

We now have enough to make an informal definition of a limit, which is actually sufficient for most of what we will do in this text.

Definition 1.3.3. Informal definition of limit.

\begin{align*} \lim_{x \to a} f(x) &= L \end{align*}

if the value of the function \(f(x)\) is sure to be arbitrarily close to \(L\) whenever the value of \(x\) is close enough to \(a\text{,}\) without 3  being exactly \(a\text{.}\)

In order to make this definition more mathematically correct, we need to make the idea of “closer and closer” more precise — we do this in Section 1.7. It should be emphasised that the formal definition and the contents of that section are optional material.

For now, let us use the above definition to examine a more substantial example.

Example 1.3.4. \(\lim_{x \to 2} \frac{x-2}{x^2+x-6}\).

Let \(f(x) = \frac{x-2}{x^2+x-6}\) and consider its limit as \(x \to 2\text{.}\)

\begin{align*} \lim_{x \to 2} \frac{x-2}{x^2+x-6} &= \text{ what?} \end{align*}

• Now if we try to compute \(f(2)\) we get \(0/0\) which is undefined. The function is not defined at that point — this is a good example of why we need limits. We have to sneak up on these places where a function is not defined (or is badly behaved).
• Very important point: the fraction \(\frac{0}{0}\) is not \(\infty\) and it is not \(1\text{,}\) it is not defined. We cannot ever divide by zero in normal arithmetic and obtain a consistent and mathematically sensible answer. If you learned otherwise in high-school, you should quickly unlearn it.

\begin{align*} \lim_{x \to 2} \frac{x-2}{x^2+x-6} &= 0.2 \end{align*}

The previous two examples are nicely behaved in that the limits we tried to compute actually exist. We now turn to two nastier examples 4  in which the limits we are interested in do not exist.

Example 1.3.5. A bad example.

Consider the following function \(f(x) = \sin( \pi /x )\text{.}\) Find the limit as \(x \to 0\) of \(f(x)\text{.}\)

We should see something interesting happening close to \(x=0\) because \(f(x)\) is undefined there. Using your favourite graph-plotting software you can see that the graph looks roughly like

How to explain this? As \(x\) gets closer and closer to zero, \(\pi/x\) becomes larger and larger (remember what the plot of \(y=1/x\) looks like). So when you take sine of that number, it oscillates faster and faster the closer you get to zero. Since the function does not approach a single number as we bring \(x\) closer and closer to zero, the limit does not exist.

We write this as

\begin{gather*} \lim_{x\to 0} \sin \left(\frac{\pi}{x}\right) \text{ does not exist} \end{gather*}

It's not very inventive notation, however it is clear. We frequently abbreviate “does not exist” to “DNE” and rewrite the above as

\begin{align*} \lim_{x\to 0} \sin \left(\frac{\pi}{x}\right) &= \text{DNE} \end{align*}

In the following example, the limit we are interested in does not exist. However the way in which things go wrong is quite different from what we just saw.

Example 1.3.6. A non-existant limit.

Consider the function

\begin{align*} f(x) &= \begin{cases} x & x \lt 2 \\ -1 & x=2 \\ x+3 & x \gt 2 \end{cases} \end{align*}

• The plot of this function looks like this

\begin{align*} \lim_{x \to 2} f(x) &= \text{DNE} \end{align*}

While the limit in the previous example does not exist, the example serves to introduce the idea of “one-sided limits”. For example, we can say that

As \(x\) moves closer and closer to two from below the function approaches 2.

and similarly

As \(x\) moves closer and closer to two from above the function approaches 5.

Definition 1.3.7. Informal definition of one-sided limits.

\begin{align*} \lim_{x \to a^-} f(x) &= K \end{align*}

when the value of \(f(x)\) gets closer and closer to \(K\) when \(x \lt a\) and \(x\) moves closer and closer to \(a\text{.}\) Since the \(x\)-values are always less than \(a\text{,}\) we say that \(x\) approaches \(a\) from below . This is also often called the left-hand limit since the \(x\)-values lie to the left of \(a\) on a sketch of the graph.

We similarly write

\begin{align*} \lim_{x \to a^+} f(x) &= L \end{align*}

when the value of \(f(x)\) gets closer and closer to \(L\) when \(x \gt a\) and \(x\) moves closer and closer to \(a\text{.}\) For similar reasons we say that \(x\) approaches \(a\) from above, and sometimes refer to this as the right-hand limit.

Note — be careful to include the superscript \(+\) and \(-\) when writing these limits. You might also see the following notations:

\begin{align*} \lim_{x \to a^+} f(x) &= \lim_{x \to a+} f(x) = \lim_{x \downarrow a} f(x) = \lim_{x \searrow a} f(x) = L & \text{right-hand limit}\\ \lim_{x \to a^-} f(x) &= \lim_{x \to a-} f(x) = \lim_{x \uparrow a} f(x) = \lim_{x \nearrow a} f(x) = L & \text{left-hand limit} \end{align*}

but please use with the notation in Definition 1.3.7 above.

Given these two similar notions of limits, when are they the same? The following theorem tell us.

Theorem 1.3.8. Limits and one sided limits.

\begin{align*} \lim_{x \to a} f(x) = L && \mbox{ if and only if } && \lim_{x \to a^-} f(x) = L \mbox{ and } \lim_{x \to a^+} f(x) = L \end{align*}

Notice that this is really two separate statements because of the “if and only if”

• If the limit of \(f(x)\) as \(x\) approaches \(a\) exists and is equal to \(L\text{,}\) then both the left-hand and right-hand limits exist and are equal to \(L\text{.}\) AND,
• If the left-hand and right-hand limits as \(x\) approaches \(a\) exist and are equal, then the limit as \(x\) approaches \(a\) exists and is equal to the one-sided limits.

That is — the limit of \(f(x)\) as \(x\) approaches \(a\) will only exist if it doesn't matter which way we approach \(a\) (either from left or right) AND if we get the same one-sided limits when we approach from left and right, then the limit exists.

We can rephrase the above by writing the contrapositives 5 of the above statements.

• If either of the left-hand and right-hand limits as \(x\) approaches \(a\) fail to exist, or if they both exist but are different, then the limit as \(x\) approaches \(a\) does not exist. AND,
• If the limit as \(x\) approaches \(a\) does not exist, then the left-hand and right-hand limits are either different or at least one of them does not exist.

Here is another limit example

Example 1.3.9. Left- and right-handed limits.

Consider the following two functions and compute their limits and one-sided limits as \(x\) approaches 1:

These are a little different from our previous examples, in that we do not have formulas, only the sketch. But we can still compute the limits.

\begin{align*} \lim_{x \to 1^-} f(x) &= 2 & \lim_{x \to 1^+} f(x) &= 2\\ \\ \end{align*}

so by the previous theorem

\begin{align*} \lim_{t \to 1^-} g(t) &= 2 & \text{and } \lim_{t \to 1^+} g(t) &= -2\\ \end{align*}

We have seen 2 ways in which a limit does not exist — in one case the function oscillated wildly, and in the other there was some sort of “jump” in the function, so that the left-hand and right-hand limits were different.

There is a third way that we must also consider. To describe this, consider the following four functions:

None of these functions are defined at \(x=a\text{,}\) nor do the limits as \(x\) approaches \(a\) exist. However we can say more than just “the limits do not exist”.

Notice that the value of function 1 can be made bigger and bigger as we bring \(x\) closer and closer to \(a\text{.}\) Similarly the value of the second function can be made arbitrarily large and negative (i.e. make it as big a negative number as we want) by bringing \(x\) closer and closer to \(a\text{.}\) Based on this observation we have the following definition.

Definition 1.3.11

\begin{gather*} \lim_{x \to a} f(x) = + \infty \end{gather*}

when the value of the function \(f(x)\) becomes arbitrarily large and positive as \(x\) gets closer and closer to \(a\text{,}\) without being exactly \(a\text{.}\)

Similarly, we write

\begin{gather*} \lim_{x \to a} f(x) = - \infty \end{gather*}

when the value of the function \(f(x)\) becomes arbitrarily large and negative as \(x\) gets closer and closer to \(a\text{,}\) without being exactly \(a\text{.}\)

A good examples of the above is

\begin{align*} \lim_{x \to 0} \frac{1}{x^2} &= +\infty & \lim_{x \to 0} -\frac{1}{x^2} &= -\infty \end{align*}

IMPORTANT POINT: Please do not think of “\(+\infty\)” and “\(-\infty\)” in these statements as numbers. You should think of \(\lim\limits_{x\rightarrow a} f(x) = +\infty\) and \(\lim\limits_{x\rightarrow a} f(x) = -\infty\) as special cases of \(\lim\limits_{x\rightarrow a} f(x) = \text{DNE}\text{.}\) The statement

\begin{gather*} \lim_{x \to a} f(x) = +\infty \end{gather*}

does not say “the limit of \(f(x)\) as \(x\) approaches \(a\) is positive infinity”. It says “the function \(f(x)\) becomes arbitrarily large as \(x\) approaches \(a\)”. These are different statements; remember that \(\infty\) is not a number 6  .

Now consider functions 3 and 4 in Figure 1.3.10. Here we can make the value of the function as big and positive as we want (for function 3) or as big and negative as we want (for function 4) but only when \(x\) approaches \(a\) from one side. With this in mind we can construct similar notation and a similar definition:

Definition 1.3.12.

\begin{gather*} \lim_{x \to a^+} f(x) = + \infty \end{gather*}

when the value of the function \(f(x)\) becomes arbitrarily large and positive as \(x\) gets closer and closer to \(a\) from above (equivalently — from the right), without being exactly \(a\text{.}\)

\begin{gather*} \lim_{x \to a^+} f(x) = - \infty \end{gather*}

when the value of the function \(f(x)\) becomes arbitrarily large and negative as \(x\) gets closer and closer to \(a\) from above (equivalently — from the right), without being exactly \(a\text{.}\)

The notation

\begin{align*} \lim_{x \to a^-} f(x) &= + \infty & \lim_{x \to a^-} f(x) &= - \infty \end{align*}

has a similar meaning except that limits are approached from below / from the left.

So for function 3 we have

\begin{align*} \lim_{x\to a^-} f(x) &= +\infty & \lim_{x\to a^+} f(x) &= \text{some positive number} \end{align*}

and for function 4

\begin{align*} \lim_{x\to a^-} f(x) &= \text{some positive number} & \lim_{x\to a^+} f(x) &= -\infty \end{align*}

More examples:

Example 1.3.13. \(\lim_{x \to \pi} \frac{1}{\sin(x)}\).

\begin{align*} g(x) &= \frac{1}{\sin(x)} \end{align*}

Find the one-sided limits of this function as \(x \to \pi\text{.}\)

Probably the easiest way to do this is to first plot the graph of \(\sin(x)\) and \(1/x\) and then think carefully about the one-sided limits:

So as \(x \to \pi\) from the left, \(\sin(x) \to 0\) from above, and so \(1/\sin(x) \to +\infty\text{.}\)

• By very similar reasoning, as \(x \to \pi\) from the right, \(\sin(x)\) is a small negative number that gets closer and closer to zero. So as \(x \to \pi\) from the right, \(\sin(x) \to 0\) through negative numbers (i.e. from below) and so \(1/\sin(x)\) to \(-\infty\text{.}\)

\begin{align*} \lim_{x \to \pi^-} \frac{1}{\sin(x)} &= +\infty & \lim_{x \to \pi^+} \frac{1}{\sin(x)} &= -\infty \end{align*}

Again, we can make Definitions 1.3.11 and 1.3.12 into mathematically precise formal definitions using techniques very similar to those in the optional Section 1.7. This is not strictly necessary for this course.

Up to this point we explored limits by sketching graphs or plugging values into a calculator. This was done to help build intuition, but it is not really the basis of a systematic method for computing limits. We have also avoided more formal approaches 7  since we do not have time in the course to go into that level of detail and (arguably) we don't need that detail to achieve the aims of the course. Thankfully we can develop a more systematic approach based on the idea of building up complicated limits from simpler ones by examining how limits interact with the basic operations of arithmetic.

Exercise \(\PageIndex{1}\)

Given the function shown below, evaluate the following:

• \(\displaystyle \lim_{x \rightarrow -2} f(x)\)
• \(\displaystyle \lim_{x \rightarrow 0}f(x)\)
• \(\displaystyle \lim_{x \rightarrow 2}f(x)\)

Exercise \(\PageIndex{2}\)

Given the function shown below, evaluate \(\displaystyle \lim_{x \rightarrow 0} f(x)\text{.}\)

Exercise \(\PageIndex{3}\)

Given the function shown below, evaluate:

• \(\displaystyle \lim_{x \rightarrow -1^{-}} f(x)\)
• \(\displaystyle \lim_{x \rightarrow -1^{+}} f(x)\)
• \(\displaystyle \lim_{x \rightarrow -1} f(x)\)
• \(\displaystyle \lim_{x \rightarrow -2^{+}} f(x)\)
• \(\displaystyle \lim_{x \rightarrow 2^{-}} f(x)\)

Exercise \(\PageIndex{4}\)

Draw a curve \(y=f(x)\) with \(\displaystyle\lim_{x \rightarrow 3}f(x)=f(3)=10\text{.}\)

Exercise \(\PageIndex{5}\)

Draw a curve \(y=f(x)\) with \(\displaystyle\lim_{x \rightarrow 3}f(x)=10\) and \(f(3)=0\text{.}\)

Exercise \(\PageIndex{6}\)

Suppose \(\displaystyle\lim_{x \rightarrow 3} f(x)=10\text{.}\) True or false: \(f(3)=10\text{.}\)

Exercise \(\PageIndex{7}\)

Suppose \(f(3)=10\text{.}\) True or false: \(\displaystyle\lim_{x \rightarrow 3} f(x)=10\text{.}\)

Exercise \(\PageIndex{8}\)

Suppose \(f(x)\) is a function defined on all real numbers, and \(\displaystyle\lim_{x \rightarrow -2} f(x)=16\text{.}\) What is \(\displaystyle\lim_{x \rightarrow -2^-} f(x)\text{?}\)

Exercise \(\PageIndex{9}\)

Suppose \(f(x)\) is a function defined on all real numbers, and \(\displaystyle\lim_{x \rightarrow -2^-} f(x)=16\text{.}\) What is \(\displaystyle\lim_{x \rightarrow -2} f(x)\text{?}\)

In Questions 1.3.2.10 through 1.3.2.17, evaluate the given limits. If you aren't sure where to begin, it's nice to start by drawing the function.

Exercise \(\PageIndex{10}\)

\(\displaystyle\lim_{t \rightarrow 0} \sin t\)

Exercise \(\PageIndex{11}\)

\(\displaystyle\lim_{x \rightarrow 0^+} \log x\)

Exercise \(\PageIndex{12}\)

\(\displaystyle\lim_{y \rightarrow 3} y^2\)

Exercise \(\PageIndex{13}\)

\(\displaystyle\lim_{x \rightarrow 0^-} \dfrac{1}{x}\)

Exercise \(\PageIndex{14}\)

\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x}\)

Exercise \(\PageIndex{15}\)

\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x^2}\)

Exercise \(\PageIndex{16}\)

\(\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}\)

Exercise \(\PageIndex{17}\)

\(\displaystyle\lim_{x \rightarrow 3} f(x)\text{,}\) where \(f(x)=\left\{ \begin{array}{ll} \sin x&x\leq 2.9\\ x^2&x \gt 2.9 \end{array} \right.\text{.}\)

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1.3: Limits

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Mathematical Expression Editor

In this section we use the graph of a function to find limits.

Finding Limits Graphically

We now consider several examples of limits of functions presented graphically.

In general, when the one-sided limits are different then the corresponding two-sided limit does not exist.

Here is a problem that encompasses all of the ideas in this section.

In the next few problems, we will reverse the situation. Information about the limits and values of a function are given and you are asked to construct the graph of the function. Many answers are possible, but if your answer is correct, then your graph will have all of the properties listed.