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Active Calculus

Section 2.8 Using Derivatives to Evaluate Limits

Because differential calculus is based on the definition of the derivative, and the definition of the derivative involves a limit, there is a sense in which all of calculus rests on limits. In addition, the limit involved in the definition of the derivative always generates the indeterminate form 00. If f is a differentiable function, then in the definition
f(x)=limh0f(x+h)f(x)h,
not only does h0 in the denominator, but also (f(x+h)f(x))0 in the numerator, since f is continuous. Remember, saying that a limit has an indeterminate form only means that we don’t yet know its value and have more work to do: indeed, limits of the form 00 can take on any value, as is evidenced by evaluating f(x) for varying values of x for a function such as f(x)=x2.
We have learned many techniques for evaluating the limits that result from the derivative definition, including a large number of shortcut rules. In this section, we turn the situation upside-down: instead of using limits to evaluate derivatives, we explore how to use derivatives to evaluate certain limits.

Preview Activity 2.8.1.

Let h be the function given by h(x)=x5+x2x21.

(a) Parts a and b.

What is the domain of h?
  • A. All real numbers except 1 and 1
  • B. All real numbers except 0
  • C. All real numbers except 0, 1, and 1
  • D. All real numbers
  • E. All real numbers except 0 and 1
  • F. All real numbers except 1
  • G. None of these
Choose the reason why limx1x5+x2x21 results in an indeterminate form.
  • A. The denominator approaches 0 as x1.
  • B. The limit is 0.
  • C. We can’t evaluate the limit.
  • D. The numerator approaches 0 as x1.
  • E. The limit does not exist.
  • F. Both the numerator and denominator approach 0 as x1.
  • G. None of these.

(b) Parts c and d.

Next, we will investigate the behavior of both the numerator and denominator of h(x) near the point where x=1. Let f(x)=x5+x2 and g(x)=x21.
Find the local linearizations (or tangent line approximations) of f and g at a=1, and call these functions Lf(x) and Lg(x), respectively.
Lf(x)=
Lg(x)=
Because the local linearizations approximate the functions near a=1, it makes sense that h(x)Lf(x)Lg(x) for x near a=1.

(c) Parts e and f.

Evaluate the limit:
limx1Lf(x)Lg(x)=
We investigate the function h(x) numerically near a=1 and compare to what we got for limx1Lf(x)Lg(x) above.
Calculate h(1.1)=
Calculate h(1.01)=
Calculate h(0.99)=
Calculate h(0.9)=

Subsection 2.8.1 Using derivatives to evaluate indeterminate limits of the form 00.

Figure 2.8.1. At left, the graphs of f and g near the value a, along with their tangent line approximations Lf and Lg at x=a. At right, zooming in on the point a and the four graphs.
The idea demonstrated in Preview Activity 2.8.1 — that we can evaluate an indeterminate limit of the form 00 by replacing each of the numerator and denominator with their local linearizations at the point of interest — can be generalized in a way that enables us to evaluate a wide range of limits. We have a function h(x) that can be written as a quotient h(x)=f(x)g(x), where f and g are both differentiable at x=a and for which f(a)=g(a)=0. We would like to evaluate the indeterminate limit given by limxah(x). Figure 2.8.1 illustrates the situation. We see that both f and g have an x-intercept at x=a. Their respective tangent line approximations Lf and Lg at x=a are also shown in the figure. We can take advantage of the fact that a function and its tangent line approximation become indistinguishable as xa.
First, let’s recall that Lf(x)=f(a)(xa)+f(a) and Lg(x)=g(a)(xa)+g(a). Because x is getting arbitrarily close to a when we take the limit, we can replace f with Lf and replace g with Lg, and thus we observe that
limxaf(x)g(x)=(limxaLf(x)Lg(x)=(limxaf(a)(xa)+f(a)g(a)(xa)+g(a).
Next, we remember that both f(a)=0 and g(a)=0, which is precisely what makes the original limit indeterminate. Substituting these values for f(a) and g(a) in the limit above, we now have
limxaf(x)g(x)=(limxaf(a)(xa)g(a)(xa)=(limxaf(a)g(a),
where the latter equality holds because xaxa=1 when x is approaching (but not equal to) a. Finally, we note that f(a)g(a) is constant with respect to x, and thus
limxaf(x)g(x)=f(a)g(a).
This result holds as long as g(a) is not equal to zero. The formal name of the result is L’Hôpital’s Rule.

L’Hôpital’s Rule.

Let f and g be differentiable at x=a, and suppose that f(a)=g(a)=0 and that g(a)0. Then limxaf(x)g(x)=f(a)g(a).
In practice, we typically work with a slightly more general version of L’Hôpital’s Rule, which states that (under the identical assumptions as the boxed rule above and the extra assumption that g is continuous at x=a)
limxaf(x)g(x)=limxaf(x)g(x),
provided the righthand limit exists. This form reflects the basic idea of L’Hôpital’s Rule: if f(x)g(x) produces an indeterminate limit of form 00 as xa, that limit is equivalent to the limit of the quotient of the two functions’ derivatives, f(x)g(x).
For example, if we consider the limit from Preview Activity 2.8.1,
limx1x5+x2x21,
by L’Hôpital’s Rule we have that
limx1x5+x2x21=limx15x4+12x=62=3.
By replacing the numerator and denominator with their respective derivatives, we often replace an indeterminate limit with one whose value we can easily determine.

Activity 2.8.2.

Evaluate each of the following limits. If you use L’Hôpital’s Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.
  1. limx0ln(1+x)x
  2. limxπcos(x)x
  3. limx12ln(x)1ex1
  4. limx0sin(x)xcos(2x)1
While L’Hôpital’s Rule can be applied in an entirely algebraic way, it is important to remember that the justification of the rule is graphical: the main idea is that the slopes of the tangent lines to f and g at x=a determine the value of the limit of f(x)g(x) as xa.
Figure 2.8.2. Two functions f and g that satisfy L’Hôpital’s Rule.
We see this in Figure 2.8.2, where we can see from the grid that f(a)=2 and g(a)=1, hence by L’Hôpital’s Rule,
limxaf(x)g(x)=f(a)g(a)=21=2.
It’s not the fact that f and g both approach zero that matters most, but rather the rate at which each approaches zero that determines the value of the limit. This is a good way to remember what L’Hôpital’s Rule says: if f(a)=g(a)=0, the the limit of f(x)g(x) as xa is given by the ratio of the slopes of f and g at x=a.

Activity 2.8.3.

In this activity, we reason graphically from the following figure to evaluate limits of ratios of functions about which some information is known.
Figure 2.8.3. Three graphs referenced in the questions of Activity 2.8.3.
  1. Use the left-hand graph to determine the values of f(2), f(2), g(2), and g(2). Then, evaluate limx2f(x)g(x).
  2. Use the middle graph to find p(2), p(2), q(2), and q(2). Then, determine the value of limx2p(x)q(x).
  3. Assume that r and s are functions whose for which r(2)0 and s(2)0 Use the right-hand graph to compute r(2), r(2), s(2), s(2). Explain why you cannot determine the exact value of limx2r(x)s(x) without further information being provided, but that you can determine the sign of limx2r(x)s(x). In addition, state what the sign of the limit will be, with justification.

Subsection 2.8.2 Limits involving

The concept of infinity, denoted , arises naturally in calculus, as it does in much of mathematics. It is important to note from the outset that is a concept, but not a number itself. Indeed, the notion of naturally invokes the idea of limits. Consider, for example, the function f(x)=1x, whose graph is pictured in Figure 2.8.4.
We note that x=0 is not in the domain of f, so we may naturally wonder what happens as x0. As x0+, we observe that f(x) increases without bound. That is, we can make the value of f(x) as large as we like by taking x closer and closer (but not equal) to 0, while keeping x>0. This is a good way to think about what infinity represents: a quantity is tending to infinity if there is no single number that the quantity is always less than.
Figure 2.8.4. The graph of f(x)=1x.
Recall that the statement limxaf(x)=L, means that we can make f(x) as close to L as we’d like by taking x sufficiently close (but not equal) to a. We now expand this notation and language to include the possibility that either L or a can be . For instance, for f(x)=1x, we now write
limx0+1x=,
by which we mean that we can make 1x as large as we like by taking x sufficiently close (but not equal) to 0. In a similar way, we write
limx1x=0,
since we can make 1x as close to 0 as we’d like by taking x sufficiently large (i.e., by letting x increase without bound).
In general, the notation limxaf(x)= means that we can make f(x) as large as we like by taking x sufficiently close (but not equal) to a, and the notation limxf(x)=L means that we can make f(x) as close to L as we like by taking x sufficiently large. This notation also applies to left- and right-hand limits, and to limits involving . For example, returning to Figure 2.8.4 and f(x)=1x, we can say that
limx01x=  and  limx1x=0.
Finally, we write
limxf(x)=
if we can make the value of f(x) as large as we’d like by taking x sufficiently large. For example,
limxx2=.
Limits involving infinity identify vertical and horizontal asymptotes of a function. If limxaf(x)=, then x=a is a vertical asymptote of f, while if limxf(x)=L, then y=L is a horizontal asymptote of f. Similar statements can be made using , and with left- and right-hand limits as xa or xa+.
In precalculus classes, it is common to study the end behavior of certain families of functions, by which we mean the behavior of a function as x and as x. Here we briefly examine some familiar functions and note the values of several limits involving .
Figure 2.8.5. Graphs of some familiar functions whose end behavior as x± is known. In the middle graph, f(x)=x316x and g(x)=x416x28.
For the natural exponential function ex, we note that limxex= and limxex=0. For the exponential decay function ex, these limits are reversed, with limxex=0 and limxex=. Turning to the natural logarithm function, we have limx0+ln(x)= and limxln(x)=. While both ex and ln(x) grow without bound as x, the exponential function does so much more quickly than the logarithm function does. We’ll soon use limits to quantify what we mean by “quickly.”
For polynomial functions of the form
p(x)=anxn+an1xn1+a1x+a0,
the end behavior depends on the sign of an and whether the highest power n is even or odd. If n is even and an is positive, then limxp(x)= and limxp(x)=, as in the plot of g in Figure 2.8.5. If instead an is negative, then limxp(x)= and limxp(x)=. In the situation where n is odd, then either limxp(x)= and limxp(x)= (which occurs when an is positive, as in the graph of f in Figure 2.8.5), or limxp(x)= and limxp(x)= (when an is negative).
A function can fail to have a limit as x. For example, consider the plot of the sine function at right in Figure 2.8.5. Because the function continues oscillating between 1 and 1 as x, we say that limxsin(x) does not exist.
Finally, it is straightforward to analyze the behavior of any rational function as x.

Example 2.8.6.

Determine the limit of the function
q(x)=3x24x+57x2+9x10
as x.
Note that both (3x24x+5) as x and (7x2+9x10) as x. Here we say that limxq(x) has indeterminate form . We can determine the value of this limit through a standard algebraic approach. Multiplying the numerator and denominator each by 1x2, we find that
limxq(x)=(limx3x24x+57x2+9x101x21x2=(limx341x+51x27+91x101x2=37
since 1x20 and 1x0 as x. This shows that the rational function q has a horizontal asymptote at y=37. A similar approach can be used to determine the limit of any rational function as x.
But how should we handle a limit such as
limxx2ex?
Here, both x2 and ex, but there is not an obvious algebraic approach that enables us to find the limit’s value. Fortunately, it turns out that L’Hôpital’s Rule extends to cases involving infinity.

L’Hôpital’s Rule ().

If f and g are differentiable and both approach zero or both approach ± as xa (where a is allowed to be ) , then
limxaf(x)g(x)=limxaf(x)g(x),
provided the righthand limit exists.
(To be technically correct, we need to add the additional hypothesis that g(x)0 on an open interval that contains a or in every neighborhood of infinity if a is ; this is almost always met in practice.)
To evaluate limxx2ex, we can apply L’Hôpital’s Rule, since both x2 and ex. Doing so, it follows that
limxx2ex=limx2xex.
This updated limit is still indeterminate and of the form , but it is simpler since 2x has replaced x2. Hence, we can apply L’Hôpital’s Rule again, and find that
limxx2ex=limx2xex=limx2ex.
Now, since 2 is constant and ex as x, it follows that 2ex0 as x, which shows that
limxx2ex=0.

Activity 2.8.4.

Evaluate each of the following limits. If you use L’Hôpital’s Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.
  1. limxxln(x)
  2. limxex+x2ex+x2
  3. limx0+ln(x)1x
  4. limxπ2tan(x)xπ2
  5. limxxex
To evaluate the limit of a quotient of two functions f(x)g(x) that results in an indeterminate form of , in essence we are asking which function is growing faster without bound. We say that the function g dominates the function f as x provided that
limxf(x)g(x)=0,
whereas f dominates g provided that limxf(x)g(x)=. Finally, if the value of limxf(x)g(x) is finite and nonzero, we say that f and g grow at the same rate. For example, we saw that limxx2ex=0, so ex dominates x2, while limx3x24x+57x2+9x10=37, so f(x)=3x24x+5 and g(x)=7x2+9x10 grow at the same rate.

Subsection 2.8.3 Summary

  • Derivatives can be used to help us evaluate indeterminate limits of the form 00 through L’Hôpital’s Rule, by replacing the functions in the numerator and denominator with their tangent line approximations. In particular, if f(a)=g(a)=0 and f and g are differentiable at a, L’Hôpital’s Rule tells us that
    limxaf(x)g(x)=limxaf(x)g(x).
  • When we write x, this means that x is increasing without bound. Thus, limxf(x)=L means that we can make f(x) as close to L as we like by choosing x to be sufficiently large. Similarly, limxaf(x)=, means that we can make f(x) as large as we like by choosing x sufficiently close to a.
  • A version of L’Hôpital’s Rule also helps us evaluate indeterminate limits of the form . If f and g are differentiable and both approach zero or both approach ± as xa (where a is allowed to be ), then
    limxaf(x)g(x)=limxaf(x)g(x).

Exercises 2.8.4 Exercises

1.

Based on your knowledge of the behavior of the numerator and denominator, predict the value of the limit limx0+sinxx1/3
the limit is
Find the limit using l’Hopital’s rule:
limx0+sinxx1/3=limx0+ [ / ] =

2.

Evaluate the following limit using L’Hospital’s rule where appropriate.
limx0sin(7x)tan(11x)
Answer:

3.

Find the limit. Use l’Hospital’s Rule where appropriate.
limxx2ex
Limit:

4.

Find the limit. Use l’Hospital’s Rule where appropriate.
limx8(lnx)3x
Limit:

5.

If f is continuous, f(7)=0, and f(7)=13, evaluate
limx0f(7+3x)+f(7+5x)x.
Limit:

6.

Evaluate the limit
limx10+4x2(3+2x)
Enter I for , -I for , and DNE if the limit does not exist.
Limit =

7.

A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 L/min. The concentration of salt after t minutes can be shown to be given by:
C(t)=30t200+t
What happens to the concentration as t?
limtC(t)= g/L

8.

Evaluate the limit
limx(x2+2x27)
Answer:

9.

The functions f and g and their tangent lines at (4,0) are shown in the figure below.
f is shown in blue, g in black, and the tangent line to f is y=0.8(x4), and is graphed in red, and the tangent line to g is y=0.4(x4), and is graphed in green.
Find the limit
limx4f(x)g(x)=

10.

For the figures below, determine the nature of limxaf(x)g(x), if f(x) is shown as the blue curve and g(x) as the black curve.
limxaf(x)g(x)=
  • positive
  • negative
  • zero
  • undefined
limxaf(x)g(x)=
  • positive
  • negative
  • zero
  • undefined

11.

Let f and g be differentiable functions about which the following information is known: f(3)=g(3)=0, f(3)=g(3)=0, f(3)=2, and g(3)=1. Let a new function h be given by the rule h(x)=f(x)g(x). On the same set of axes, sketch possible graphs of f and g near x=3, and use the provided information to determine the value of
limx3h(x).
Provide explanation to support your conclusion.

12.

Find all vertical and horizontal asymptotes of the function
R(x)=3(xa)(xb)5(xa)(xc),
where a, b, and c are distinct, arbitrary constants. In addition, state all values of x for which R is not continuous. Sketch a possible graph of R, clearly labeling the values of a, b, and c.

13.

Consider the function g(x)=x2x, which is defined for all x>0. Observe that limx0+g(x) is indeterminate due to its form of 00. (Think about how we know that 0k=0 for all k>0, while b0=1 for all b0, but that neither rule can apply to 00.)
  1. Let h(x)=ln(g(x)). Explain why h(x)=2xln(x).
  2. Next, explain why it is equivalent to write h(x)=2ln(x)1x.
  3. Use L’Hôpital’s Rule and your work in (b) to compute limx0+h(x).
  4. Based on the value of limx0+h(x), determine limx0+g(x).

14.

Recall we say that function g dominates function f provided that limxf(x)=, limxg(x)=, and limxf(x)g(x)=0.
  1. Which function dominates the other: ln(x) or x?
  2. Which function dominates the other: ln(x) or xn? (n can be any positive integer)
  3. Explain why ex will dominate any polynomial function.
  4. Explain why xn will dominate ln(x) for any positive integer n.
  5. Give any example of two nonlinear functions such that neither dominates the other.
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