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APEX Calculus

Section 6.7 L’Hospital’s Rule

While this chapter is devoted to learning techniques of integration, this section is not about integration. Rather, it is concerned with a technique of evaluating certain limits that will be useful in the following section, where integration is once more discussed.
Our treatment of limits exposed us to the notion of “0/0”, an indeterminate form. If limxcf(x)=0 and limxcg(x)=0, we do not conclude that limxcf(x)/g(x) is 0/0; rather, we use 0/0 as notation to describe the fact that both the numerator and denominator approach 0. The expression 0/0 has no numeric value; other work must be done to evaluate the limit.
Other indeterminate forms exist; they are: /, 0, , 00, 1 and 0. Just as “0/0” does not mean “divide 0 by 0,” the expression “/” does not mean “divide infinity by infinity.” Instead, it means “a quantity is growing without bound and is being divided by another quantity that is growing without bound.” We cannot determine from such a statement what value, if any, results in the limit. Likewise, “0” does not mean “multiply zero by infinity.” Instead, it means “one quantity is shrinking to zero, and is being multiplied by a quantity that is growing without bound.” We cannot determine from such a description what the result of such a limit will be.
Figure 6.7.1. Video introduction to Section 6.7
This section introduces l’Hospital’s Rule, a method of resolving limits that produce the indeterminate forms 0/0 and /. We’ll also show how algebraic manipulation can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied.

Subsection 6.7.1 L’Hospital’s Rule with Indeterminate Forms 0/0 and /

We demonstrate the use of l’Hospital’s Rule in the following examples; we will often use “LHR” as an abbreviation of “l’Hospital’s Rule.”

Example 6.7.3. Using l’Hospital’s Rule.

Evaluate the following limits, using l’Hospital’s Rule as needed.
  1. limx0sin(x)x
  2. limx1x+321x
  3. limx0x21cos(x)
  4. limx2x2+x6x23x+2
Solution 1.
  1. We proved this limit is 1 in Example 1.3.13 using the Squeeze Theorem. Here we use l’Hospital’s Rule to show its power.
    limx0sin(x)x= by LHR limx0cos(x)1=1.
  2. limx1x+321x= by LHR limx112(x+3)1/21=14.
  3. limx0x21cos(x)= by LHR limx02xsin(x).
    This latter limit also evaluates to the 0/0 indeterminate form. To evaluate it, we apply l’Hospital’s Rule again.
    limx02xsin(x)= by LHR 2cos(x)=2.
    Thus limx0x21cos(x)=2.
  4. We already know how to evaluate this limit; first factor the numerator and denominator. We then have:
    limx2x2+x6x23x+2=limx2(x2)(x+3)(x2)(x1)=limx2x+3x1=5.
    We now show how to solve this using l’Hospital’s Rule.
    limx2x2+x6x23x+2= by LHR limx22x+12x3=5.
Solution 2. Video solution
Note that at each step where l’Hospital’s Rule was applied, it was needed: the initial limit returned the indeterminate form of “0/0.” If the initial limit returns, for example, 1/2, then l’Hospital’s Rule does not apply.
The following theorem extends our initial version of l’Hospital’s Rule in two ways. It allows the technique to be applied to the indeterminate form / and to limits where x approaches ±.

Example 6.7.5. Using l’Hospital’s Rule with limits involving .

Evaluate the following limits.
1.limx3x2100x+24x2+5x10002.limxexx3.
Solution 1.
  1. We can evaluate this limit already using Theorem 1.6.21; the answer is 3/4. We apply l’Hospital’s Rule to demonstrate its applicability.
    limx3x2100x+24x2+5x1000= by LHR limx6x1008x+5= by LHR limx68=34.
  2. limxexx3= by LHR limxex3x2= by LHR limxex6x= by LHR limxex6=.
    Recall that this means that the limit does not exist; as x approaches , the expression ex/x3 grows without bound. We can infer from this that ex grows “faster” than x3; as x gets large, ex is far larger than x3. (This has important implications in computing when considering efficiency of algorithms.)
Solution 2. Video solution

Subsection 6.7.2 Indeterminate Forms 0 and

L’Hospital’s Rule can only be applied to ratios of functions. When faced with an indeterminate form such as 0 or , we can sometimes apply algebra to rewrite the limit so that l’Hospital’s Rule can be applied. We demonstrate the general idea in the next example.

Example 6.7.6. Applying l’Hospital’s Rule to other indeterminate forms.

Evaluate the following limits.
  1. limx0+xe1/x
  2. limx0xe1/x
  3. limxln(x+1)ln(x)
  4. limxx2ex
Solution 1.
  1. As x0+, x0 and e1/x. Thus we have the indeterminate form 0. We rewrite the expression xe1/x as e1/x1/x; now, as x0+, we get the indeterminate form / to which l’Hospital’s Rule can be applied.
    limx0+xe1/x=limx0+e1/x1/x= by LHR limx0+(1/x2)e1/x1/x2=limx0+e1/x=.
    Interpretation: e1/x grows “faster” than x shrinks to zero, meaning their product grows without bound.
  2. As x0, x0 and e1/xe0. The the limit evaluates to 00 which is not an indeterminate form. We conclude then that
    limx0xe1/x=0.
  3. This limit initially evaluates to the indeterminate form . By applying a logarithmic rule, we can rewrite the limit as
    limxln(x+1)ln(x)=limxln(x+1x).
    As x, the argument of the ln term approaches /, to which we can apply l’Hospital’s Rule.
    limxx+1x= by LHR 11=1.
    Since x implies x+1x1, it follows that
    x implies ln(x+1x)ln(1)=0.
    Thus
    limxln(x+1)ln(x)=limxln(x+1x)=0.
    Interpretation: since this limit evaluates to 0, it means that for large x, there is essentially no difference between ln(x+1) and ln(x); their difference is essentially 0.
  4. The limit limxx2ex initially returns the indeterminate form . We can rewrite the expression by factoring out x2; x2ex=x2(1exx2). We need to evaluate how ex/x2 behaves as x:
    limxexx2= by LHR limxex2x= by LHR limxex2=.
    Thus limxx2(1ex/x2) evaluates to (), which is not an indeterminate form; rather, () evaluates to . We conclude that limxx2ex=. Interpretation: as x gets large, the difference between x2 and ex grows very large.
Solution 2. Video solution

Subsection 6.7.3 Indeterminate Forms  00, 1 and 0

When faced with an indeterminate form that involves a power, it often helps to employ the natural logarithmic function. The following Key Idea expresses the concept, which is followed by an example that demonstrates its use.

Key Idea 6.7.7. Evaluating Limits Involving Indeterminate Forms 00, 1 and 0.

If limxcln(f(x))=L, then
limxcf(x)=limxceln(f(x))=eL.

Example 6.7.8. Using l’Hospital’s Rule with indeterminate forms involving exponents.

Evaluate the following limits.
  1. limx(1+1x)x
  2. limx0+xx
Solution 1.
  1. This is equivalent to a special limit given in Theorem 1.3.17; these limits have important applications within mathematics and finance. Note that the exponent approaches while the base approaches 1, leading to the indeterminate form 1. Let f(x)=(1+1/x)x; the problem asks to evaluate limxf(x). Let’s first evaluate limxln(f(x)).
    limxln(f(x))=limxln(1+1x)x=limxxln(1+1x)=limxln(1+1x)1/x
    This produces the indeterminate form 0/0, so we apply l’Hospital’s Rule.
    =limx11+1/x(1/x2)(1/x2)=limx11+1/x=1.
    Thus limxln(f(x))=1. We return to the original limit and apply Key Idea 6.7.7.
    limx(1+1x)x=limxf(x)=limxeln(f(x))=e1=e.
  2. This limit leads to the indeterminate form 00. Let f(x)=xx and consider first limx0+ln(f(x)).
    limx0+ln(f(x))=limx0+ln(xx)=limx0+xln(x)=limx0+ln(x)1/x.
    This produces the indeterminate form / so we apply l’Hospital’s Rule.
    =limx0+1/x1/x2=limx0+x=0.
    Thus limx0+ln(f(x))=0. We return to the original limit and apply Key Idea 6.7.7.
    limx0+xx=limx0+f(x)=limx0+eln(f(x))=e0=1.
    This result is supported by the graph of f(x)=xx given in Figure 6.7.9.
    Graph of function x^x, used in this the example.
    The y axis is drawn from 0 to 4 and the x axis is drawn from 0 to 2. The function f(x)=xx is drawn as a curve opening towards the positive y axis with arrows towards the ends. The function is drawn from point (0,1) from where it dips gently then rises up slowly.
    Figure 6.7.9. A graph of f(x)=xx supporting the fact that as x0+, f(x)1
Solution 2. Video solution
Our brief revisit of limits will be rewarded in the next section where we consider improper integration. So far, we have only considered definite integrals where the bounds are finite numbers, such as 01f(x)dx. Improper integration considers integrals where one, or both, of the bounds are “infinity.” Such integrals have many uses and applications, in addition to generating ideas that are enlightening.

Exercises 6.7.4 Exercises

Terms and Concepts

1.
List the different indeterminate forms described in this section.
2.
T/F: l’Hospital’s Rule provides a faster method of computing derivatives.
  • ?
  • True
  • False
3.
l’Hospital’s Rule states that ddx[f(x)g(x)]=f(x)g(x).
  • ?
  • True
  • False
4.
Explain what the indeterminate form “1” means.
5.
Fill in the blanks: The Quotient Rule is applied to f(x)g(x) when taking ; l’Hospital’s Rule is applied when taking certain .
6.
Create (but do not evaluate!) a limit that returns “0”.
7.
Create a function f(x) such that limx1f(x) returns “00”.
8.
Create a function f(x) such that limxf(x) returns “0”.

Problems

Exercise Group.
Evaluate the given limit using l’Hospital’s rule.
9.
limx1x2+x2x1
10.
limx2x2+x6x27x+10
11.
limxπsin(x)xπ
12.
limxπ/4sin(x)cos(x)cos(2x)
13.
limx0sin(5x)x
14.
limx0sin(2x)x+2
15.
limx0sin(2x)sin(3x)
16.
limx0sin(ax)sin(bx)
17.
limx0+ex1x2
18.
limx0+exx1x2
19.
limx0+xsin(x)x3x2
26.
limx3x35x2+3x+9x37x2+15x9
27.
limx2x3+4x2+4xx3+7x2+16x+12
28.
limxln(x)x
29.
limxln(x2)x
30.
limxln2(x)x
31.
limx0+xln(x)
32.
limx0+xln(x)
34.
limxx3x2
35.
limxxln(x)
36.
limxxex
37.
limx0+1x2e1x
41.
limx0+(sin(x))x
Hint: use the Squeeze Theorem.
42.
limx1(1x)1x
45.
limx1+(ln(x))1x
46.
limx(1+x)1x
47.
limx(1+x2)1x
48.
limxπ/2tan(x)cos(x)
49.
limxπ/2tan(x)sin(2x)
50.
limx1+1ln(x)1x1
51.
limx3+5x29xx3
52.
limxxtan(1x)
53.
limxln3(x)x
54.
limx1x2+x2ln(x)
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