2.11 L’Hôpital’s rule
2.11 L’Hôpital’s rule
Limits and indeterminate forms
In an earlier chapter, we discussed how we can use the mathematical concept of a limit to investigate the behaviour of a function around a particular input value (limit at a point) as well as the behaviour of a function as the input values increase or decrease in value indefinitely (limit at infinity). We discussed at that time some methods we could use to evaluate the limits, but these methods were restricted to the cases when the behaviour of the function could be evaluated close to the given input values and this evaluation used to project the function values as we get arbitrarily close to the given input. However, there are many cases when evaluation “close by” will not be helpful to make projections about the behaviour in the given neighbourhood. One of those cases is the so called case of indeterminate form. What is meant by that is that one can not determine in a straightforward way what the resulting limit is and that happens in particular when we have the following cases:
- [latex]\lim\limits_{x\to a} \dfrac{\overbrace{f(x)}^{\to 0}}{\underbrace{g(x)}_{\to 0}}[/latex], sometimes abbreviated by [latex]"\frac{0}{0}"[/latex]
- [latex]\lim\limits_{x\to a} \dfrac{\overbrace{f(x)}^{\to \pm \infty}}{\underbrace{g(x)}_{\to \pm \infty}}[/latex], sometimes abbreviated by [latex]"\frac{\infty}{\infty}"[/latex]
- [latex]\lim\limits_{x\to a} \overbrace{f(x)}^{\to 0}\cdot \overbrace{g(x)}^{\to \infty}[/latex], sometimes abbreviated by [latex]"{0}\cdot{\infty}"[/latex]
- [latex]\lim\limits_{x\to a} {\underbrace{f(x)}_{\to 0}}^{\overbrace{g(x)}^{\to 0}}[/latex], sometimes abbreviated by [latex]"{0}^{0}"[/latex]
- [latex]\lim\limits_{x\to a} {\underbrace{f(x)}_{\to \infty}}^{\overbrace{g(x)}^{\to 0}}[/latex], sometimes abbreviated by [latex]"{\infty}^{0}"[/latex]
- [latex]\lim\limits_{x\to a} {\underbrace{f(x)}_{\to 1}}^{\overbrace{g(x)}^{\to \infty}}[/latex], sometimes abbreviated by [latex]"{1}^{\infty}"[/latex]
For example, recall that in rational functions (polynomial divided by a polynomial) we can use factoring and simplifying by canceling common factors when both the numerator and the denominator approach the value of zero. When evaluating limits at infinity (positive or negative), we can rewrite the function by dividing the numerator and the denominator by the largest power in the denominator and then analyzing the limits of the new numerator and the denominator. Examples of both demonstrate that both [latex]"\frac{0}{0}"[/latex] and [latex]"\frac{\infty}{\infty}"[/latex] can have infinitely many possible answers (or none at all) – thus the word indeterminate. See below for few such examples (verify each using your knowledge from the earlier section on limits):
| Indeterminate form: [latex]"\frac{0}{0}"[/latex] | Indeterminate form: [latex]"\frac{\infty}{\infty}"[/latex] |
| [latex]\lim\limits_{x\to 2} \dfrac{\overbrace{x-2}^{\to 0}}{\underbrace{x-2}_{\to 0}}=1[/latex] | [latex]\lim\limits_{x\to \infty} \dfrac{\overbrace{3x^3+x^2-4}^{\to \pm \infty}}{\underbrace{-6x^3-x+7}_{\to \pm \infty}}=-\dfrac{1}{2}[/latex] |
| [latex]\lim\limits_{x\to 2} \dfrac{\overbrace{(x-2)^2}^{\to 0}}{\underbrace{x-2}_{\to 0}}=0[/latex] | [latex]\lim\limits_{x\to \infty} \dfrac{\overbrace{3x^3+x^2-4}^{\to \pm \infty}}{\underbrace{-6x^4-x+7}_{\to \pm \infty}}=0[/latex] |
| [latex]\lim\limits_{x\to 2} \dfrac{\overbrace{x-2}^{\to 0}}{\underbrace{(x-2)^2}_{\to 0}}[/latex] does not exist | [latex]\lim\limits_{x\to \infty} \dfrac{\overbrace{3x^3+x^2-4}^{\to \pm \infty}}{\underbrace{-6x^6-x+7}_{\to \pm \infty}}=-\infty[/latex] |
Note that we have to be careful when we write [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex] because these imply division that does not have a defined result. The quotation marks are important to distinguish the form from the actual operation of division.
L’Hôpital’s rule and indeterminate forms [latex]"\frac{0}{0}"[/latex] and [latex]"\frac{\infty}{\infty}"[/latex]
The methods applicable to rational functions may not be applicable to determining the limits of other functions when faced with indeterminate forms. For example:
[latex]\lim\limits_{x\to \infty} \dfrac{\overbrace{e^x}^{\to \infty}}{\underbrace{x^2}_{\to \infty}}=?[/latex]
[latex]\lim\limits_{x\to 1} \dfrac{\overbrace{\ln x}^{\to 0}}{\underbrace{x-1}_{\to 0}}=?[/latex]
One of the methods that may work in such cases is L’Hôpital’s rule (pronounced: L’opital’s rule).
L’Hôpital’s rule
Suppose [latex]f[/latex] and [latex]g[/latex] are differentiable and [latex]g'(x)\neq0[/latex] on an open interval that contains [latex]a[/latex] (except possibly at [latex]a[/latex]). Suppose that
[latex]\lim\limits_{x\to a}f(x)=0[/latex] and [latex]\lim\limits_{x\to a}g(x)=0[/latex]
or that
[latex]\lim\limits_{x\to a}f(x)=\pm\infty[/latex] and [latex]\lim\limits_{x\to a}g(x)=\pm\infty[/latex]
Then
[latex]\lim\limits_{x\to a}\dfrac{f(x)}{g(x)}=\lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}=[/latex]
Note: L’Hospital’s Rule works even for [latex]a=\pm\infty[/latex]!
Example 1
Find the following limits, if they exist and briefly explain the meaning of the results.
a) [latex]\lim\limits_{x\to 1}\dfrac{\ln x}{x-1}[/latex]
b) [latex]\lim\limits_{x\to \infty}\dfrac{e^x}{x^2}[/latex]
Answer:
a) [latex]\lim\limits_{x\to 1}\dfrac{\ln x}{x-1}=?[/latex] Meaning?
[latex]\lim\limits_{x\to 1}\dfrac{\overbrace{\ln x}^{\to 0}}{\underbrace{x-1}_{\to 0}}\overset{L'H}{=}\lim\limits_{x\to 1}\dfrac{\frac{1}{x}}{1}=\lim\limits_{x\to 1}\dfrac{1}{x}=1[/latex]
Therefore, the function [latex]f(x)=\dfrac{\ln x}{x-1}[/latex] has a hole at [latex](1,1)[/latex] since the function is not defined at at [latex]x=1[/latex] but has a limit at [latex]x=1[/latex] which equals [latex]1[/latex].
b) [latex]\lim\limits_{x\to \infty}\dfrac{e^x}{x^2}=?[/latex] Meaning?
[latex]\lim\limits_{x\to \infty}\dfrac{\overbrace{e^x}^{\to \infty}}{\underbrace{x^2}_{\to \infty}}\overset{L'H}{=}\dfrac{\overbrace{e^x}^{\to \infty}}{\underbrace{2x}_{\to \infty}}\overset{L'H}{=}\dfrac{\overbrace{e^x}^{\to \infty}}{\underbrace{2}_{\to 2}}=\infty[/latex]
Therefore, the function [latex]f(x)=\dfrac{e^x}{x^2}[/latex] grows infinitely large as [latex]x[/latex] increases towards positive infinity. In particular, this function does not have a horizontal asymptote as input grows towards infinity.
Note that L’Hôpital’s rule will not always work.
Other indeterminate forms and L’Hôpital’s rule
We can also sometimes use L’Hôpital’s rule when facing other indeterminate forms by rewriting the original function as a quotient in a way that leads to either [latex]"\frac{0}{0}"[/latex] or [latex]"\frac{\infty}{\infty}"[/latex]. Below are few examples.
Example 2
Find [latex]\lim\limits_{x\to 0^+} x \ln x[/latex] and interpret the result.
Answer: [latex]\lim\limits_{x\to 0^+} x \ln x=?[/latex] Interpretation?
Analyze:
[latex]\lim\limits_{x\to 0^+} \overbrace{x}^{\to 0}\cdot\overbrace{\ln x}^{-\infty}[/latex]
Rewrite into the form of one of [latex]"\frac{0}{0}"[/latex] or [latex]"\frac{\infty}{\infty}"[/latex]:
[latex]\lim\limits_{x\to 0^+} x \ln x=\lim\limits_{x\to 0^+} \dfrac{\overbrace{\ln x}^{\to -\infty}}{\underbrace{x^{-1}}_{\to\infty}}\overset{L'H}{=}\lim\limits_{x\to 0^+}\dfrac{\frac{1}{x}}{-x^{-2}}=\lim\limits_{x\to 0^+}(-x)=0[/latex]
Therefore, the function [latex]f(x)=x \ln x[/latex] approaches the value of [latex]0[/latex] as [latex]x[/latex] approaches [latex]0[/latex] from the right side even though the function is not defined at [latex]0[/latex].
Example 3
Find [latex]\lim\limits_{x\to 0^+} x^x[/latex] and interpret the result.
Answer: [latex]\lim\limits_{x\to 0^+} x^x=?[/latex] Interpretation?
Analyze:
[latex]\lim\limits_{x\to 0^+} {\underbrace{x}_{\to 0}}^{\overbrace{x}^{\to 0}}[/latex]
Let [latex]y=x^x[/latex]. Then [latex]\ln y=x\ln x[/latex].
Hence, because an exponential function is a continuous function and using our result from Example 2, we have that
[latex]\lim\limits_{x\to 0^+} x^x=\lim\limits_{x\to 0^+} y=\lim\limits_{x\to 0^+} e^{\ln y}=(e)^{\lim\limits_{x\to 0^+} \ln y}=(e)^{\lim\limits_{x\to 0^+} \overbrace{x\ln x}^{\to 0}}=1[/latex]
Therefore, the function [latex]y=x^x[/latex], although not defined at [latex]x=0[/latex], approaches the value of [latex]1[/latex] as [latex]x[/latex] approaches [latex]0[/latex] from the right.
Section Exercises
Work on the following exercises. Discuss your solutions with your peers and/or course instructor.
IC4NITS Exercises 2.11 – L’Hôpital’s rule
