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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 6.8 Indeterminate Forms & l’Hospital’s Rule

Subsection 6.8.1 Before Class

https://mymedia.ou.edu/media/6.8-1/1_v4lhgyzw
Figure 57. Pre-Class Video 1

Subsubsection 6.8.1.1 l’Hospital’s Rule

Question 6.8.1.
In Calculus 1, you learned the Direct Substitution Property to compute limits. It says that if \(f\) is a polynomial or rational function, and \(a\) is in the domain of \(f\text{,}\) then \(\ds \lim_{x\to a} f(x) = f(a)\text{.}\)
  1. Consider this limit: \(\ds \lim_{x\to 2} \dfrac{x^2-4}{x-2}\text{.}\) Why can we not use the Direct Substitution Property?
  2. Rewrite the function so that you can use Direct Substitution, and then find the limit.
Solution.
  1. You get division by zero if you were to plug in 2.
  2. \(\displaystyle \ds \lim_{x\to 2} \dfrac{x^2-4}{x-2} = \lim_{x\to 2} x+2\)
Definition 6.8.2. Indeterminate Form (Type \(0/0\) and \(\infty/\infty\).
An indeterminate form is a limit of the form \(\ds \lim_{x\to a} \dfrac{f(x)}{g(x)}\text{,}\) where either \(f(x),g(x)\to 0\) or \(f(x),g(x)\to \pm\infty\)
Example 6.8.3.
Each limit below is an indeterminate form. Classify it as Type \(0/0\) or Type \(\infty/\infty\)
  1. \(\displaystyle \ds\lim_{x\to 1} \dfrac{\ln x}{x - 1}\)
  2. \(\displaystyle \ds \lim_{x\to \pi/2^-} \dfrac{\sec x}{1+\tan x}\)
  3. \(\displaystyle \ds \lim_{x\to \infty} \dfrac{e^x}{x^2}\)
Solution.
  1. \(\displaystyle 0/0\)
  2. \(\displaystyle \infty/\infty\)
  3. \(\displaystyle \infty/\infty\)
l’Hospital’s Rule.
Let \(f\) and \(g\) be differentiable functions such that \(g'(x)\neq 0\) on an open interval \(I\) that contains \(a\) (except possibly at \(a\)). Suppose that the limit creates an indeterminate form of Type \(0/0\) or \(\infty/\infty\text{.}\) Then,
\begin{equation*} \lim_{x\to a} \dfrac{f(x)}{g(x)} = \lim_{x\to a} \dfrac{f'(x)}{g'(x)} \end{equation*}
if the limit on the right side exists, or is \(\pm \infty\text{.}\)
Example 6.8.4.
Use l’Hospital’s Rule to compute the limits:
  1. \(\displaystyle \ds\lim_{x\to 1} \dfrac{\ln x}{x - 1}\)
  2. \(\displaystyle \ds \lim_{x\to \pi/2^-} \dfrac{\sec x}{1+\tan x}\)
  3. \(\displaystyle \ds \lim_{x\to \infty} \dfrac{e^x}{x^2}\)
Solution.
  1. \(\displaystyle \ds\lim_{x\to 1} \dfrac{\ln x}{x - 1} = 1\)
  2. \(\displaystyle \ds \lim_{x\to \pi/2^-} \dfrac{\sec x}{1+\tan x}=1\)
  3. \(\displaystyle \ds \lim_{x\to \infty} \dfrac{e^x}{x^2}=\infty\)
Example 6.8.5.
Show that \(\ds \lim_{x \to 0} \dfrac{\tan x - x}{x^3} = \dfrac{1}{3}\)
Solution.
The expression \(\dfrac{\tan x - x}{x^3}\) is a \(0/0\) indeterminate form, so we may use l’Hospital’s Rule. Applying it gives
\begin{equation*} \ds \lim_{x \to 0} \dfrac{\tan x - x}{x^3} \overset{LH}{=} \lim_{x\to 0} \dfrac{\sec^2x - 1}{3x^2} \end{equation*}
Attempting to evaluate gives another \(0/0\) indeterminate form, so applying l’Hospital’s Rule again gives
\begin{equation*} \ds \lim_{x\to 0} \dfrac{\sec^2x -1}{3x^2} \overset{LH}{=} \lim_{x\to 0} \dfrac{2\sec^2x\tan x}{6x} \end{equation*}
Again we have a \(0/0\) indeterminate form, so apply the Rule a third time:
\begin{equation*} \ds \lim_{x\to 0} \dfrac{2\sec^2x\tan x}{6x} \overset{LH}{=} \lim_{x\to 0} \dfrac{4\sec^2x\tan x + 2\sec^4x}{6} \end{equation*}
Now we can resolve the limit as \(\dfrac{0 + 2}{6} = \dfrac{1}{3}\)

Subsection 6.8.2 Pre-Class Activities

Example 6.8.6.

Compute these limits; you may or may not need l’Hospital’s Rule.
  1. \(\displaystyle \ds \lim_{x\to -2} \dfrac{x^3+8}{x+2}\)
  2. \(\displaystyle \ds \lim_{x\to (\pi/2)^+} \dfrac{\cos x}{1-\sin x}\)
  3. \(\displaystyle \ds \lim_{x\to \pi^-} \dfrac{\sin x}{1-\cos x}\)
  4. \(\displaystyle \ds \lim_{u\to \infty} \dfrac{e^{u/10}}{u^3}\)
Solution.
  1. \(\ds \lim_{x\to -2} \dfrac{x^3+8}{x+2} = -12\text{;}\) l’Hospital’s Rule is not needed, but may be used.
  2. \(\ds \lim_{x\to (\pi/2)^+} \dfrac{\cos x}{1-\sin x}= -\infty\text{;}\) l’Hospital’s Rule is required here.
  3. \(\ds \lim_{x\to \pi^-} \dfrac{\sin x}{1-\cos x} = 0\text{;}\) l’Hospital’s Rule may not be used here.
  4. \(\ds \lim_{u\to \infty} \dfrac{e^{u/10}}{u^3} = \infty\text{;}\) l’Hospital’s Rule is required here.

Subsection 6.8.3 In Class

Subsubsection 6.8.3.1 Other Indeterminate Forms

Definition 6.8.7. Indeterminate Form (Type \(0\cdot \infty\)).
An indeterminate form of type \(0\cdot \infty\) is a limit of the form \(\ds \lim_{x\to a} f(x)g(x)\text{,}\) where \(\ds \lim_{x\to a} f(x) = 0\) and \(\ds \lim_{x\to a}g(x) = \pm \infty\)
Handling Type \(0\cdot \infty\) Indeterminate Forms.
Rewrite \(f(x)g(x)\) as \(\dfrac{f(x)}{1/g(x)}\) or \(\dfrac{g(x)}{1/f(x)}\)
Example 6.8.8.
Find \(\ds\lim_{x\to 0^+} x\ln x\)
Solution.
\(\ds\lim_{x\to 0^+} x\ln x = 0\)
Example 6.8.9.
Evaluate \(\ds\lim_{x\to \infty} x\sin\lrpar{\dfrac{\pi}{x}}\)
Solution.
\(\ds\lim_{x\to \infty} x\sin\lrpar{\dfrac{\pi}{x}} = \pi\)
Example 6.8.10.
Show that \(\ds \lim_{x\to (\pi/2)^-} \cos x\sec 5x = \dfrac{1}{5}\)
Solution.
We can verify that this is an indeterminate form of Type \(0\cdot \infty\text{.}\) Rewrite the limit as:
\begin{equation*} \ds \lim_{x\to (\pi/2)^-} \cos x\sec 5x = \lim_{x\to (\pi/2)^-} \dfrac{\cos x}{\cos 5x} \end{equation*}
Now applying l’Hospital’s Rule, we have
\begin{equation*} \ds \lim_{x\to (\pi/2)^-} \dfrac{\cos x}{\cos 5x} \overset{LH}{=} \lim_{x\to (\pi/2)^-} \dfrac{\sin x}{5\sin 5x} = \dfrac{1}{5} \end{equation*}
Definition 6.8.11. Indeterminate Form (Type \(\infty - \infty\)).
An indeterminant form of type \(\infty - \infty\) is a limit of the form \(\ds \lim_{x\to a}[f(x) - g(x)]\text{,}\) where \(\ds \lim_{x\to a}f(x) = \lim_{x\to a}g(x) = \infty\)
Example 6.8.12.
Compute \(\ds \lim_{x\to (\pi/2)^-} (\sec x - \tan x)\)
Solution.
\(\ds \lim_{x\to (\pi/2)^-} (\sec x - \tan x) = 0\)
Example 6.8.13.
Find \(\ds \lim_{x\to 1} \lrpar{\dfrac{x}{x-1} -\dfrac{1}{\ln x}}\)
Solution.
\(\ds \lim_{x\to 1} \lrpar{\dfrac{x}{x-1} -\dfrac{1}{\ln x}} = \dfrac{1}{2}\)
Definition 6.8.14. Indeterminate Form (Types \(0^0,\infty^0,1^\infty\)).
Consider the limit \(\ds \lim_{x\to a} [f(x)]^{g(x)}\text{.}\) This is an indeterminate form of:
  • Type \(0^0\) if \(\ds \lim_{x\to a}f(x) = 0\) and \(\ds \lim_{x\to a}g(x) = 0\)
  • Type \(\infty^0\) if \(\ds \lim_{x\to a}f(x) = \infty\) and \(\ds \lim_{x\to a}g(x) = 0\)
  • Type \(1^\infty\) if \(\ds \lim_{x\to a}f(x) = 1\) and \(\ds \lim_{x\to a}g(x) = \pm \infty\)
Handling Indeterminate Powers.
If \(y= f(x)^{g(x)}\text{,}\) set \(\ln y = \ln\lrpar{f(x)^{g(x)}} = g(x) \ln (f(x))\) and take the limit
Example 6.8.15.
Compute \(\ds \lim_{x\to 0^+} (1-\sin 2x)^{\cot x}\)
Solution.
\(\ds \lim_{x\to 0^+} (1-\sin 2x)^{\cot x} = e^{-2}\)
Example 6.8.16.
Find \(\ds \lim_{x\to 0^+} x^x\)
Solution.
\(\ds \lim_{x\to 0^+} x^x=1\)

Subsubsection 6.8.3.2 Examples

Example 6.8.17.
Find the limits:
  1. \(\displaystyle \ds \lim_{x\to \infty} \dfrac{(\ln x)^2}{x}\)
  2. \(\displaystyle \ds \lim_{x\to 1^+} \ln x\tan\lrpar{\dfrac{\pi x}{2}}\)
  3. \(\displaystyle \ds \lim_{x\to 0^+} \lrpar{\dfrac{1}{x} - \dfrac{1}{e^x-1}}\)
  4. \(\displaystyle \ds \lim_{x\to 0^+} \lrpar{\dfrac{1}{x} -\dfrac{1}{\inv{\tan}(x)}}\)
  5. \(\displaystyle \ds \lim_{x\to 0^+} (\tan 2x)^x\)
  6. \(\displaystyle \ds \lim_{x\to \infty} x^{e^{-x}}\)
Solution.
  1. \(\displaystyle \ds \lim_{x\to \infty} \dfrac{(\ln x)^2}{x} = 0\)
  2. \(\displaystyle \ds \lim_{x\to 1^+} \ln x\tan\lrpar{\dfrac{\pi x}{2}} = -\dfrac{2}{\pi}\)
  3. \(\displaystyle \ds \lim_{x\to 0^+} \lrpar{\dfrac{1}{x} - \dfrac{1}{e^x-1}} = \dfrac{1}{2}\)
  4. \(\displaystyle \ds \lim_{x\to 0^+} \lrpar{\dfrac{1}{x} -\dfrac{1}{\inv{\tan}(x)}}= 0\)
  5. \(\displaystyle \ds \lim_{x\to 0^+} (\tan 2x)^x = 1\)
  6. \(\displaystyle \ds \lim_{x\to \infty} x^{e^{-x}} = 1\)
Example 6.8.18.
The equation of the tangent line to \(f(x)\) at the point \((2,0)\) is \(1.5x-3\text{,}\) and the equation of the tangent line to \(g(x)\) at \((2,0)\) is \(2-x\text{.}\) What is \(\ds \lim_{x\to 2} \dfrac{f(x)}{g(x)}\text{?}\)
Solution.
\(\ds \lim_{x\to 2} \dfrac{f(x)}{g(x)} = -1.5\)
Example 6.8.19.
If \(f'\) is continuous, \(f(2) = 0\text{,}\) and \(f'(2) = 7\text{,}\) then find \(\ds \lim_{x\to 0} \dfrac{f(2+3x) + f(2+5x)}{x}\)
Solution.
\(\ds \lim_{x\to 0} \dfrac{f(2+3x) + f(2+5x)}{x} = 56\)
Example 6.8.20.
For what values of \(a\) and \(b\) is the equation \(\ds \lim_{x\to 0} \lrpar{\dfrac{\sin 2x}{x^3} + a + \dfrac{b}{x^2}} = 0\) true?
Solution.
\(b = -2\) and \(a = -\dfrac{4}{3}\)

Subsection 6.8.4 After Class Activities

Example 6.8.21.

Compute the limits:
  1. \(\displaystyle \ds \lim_{x\to 0} \dfrac{\inv{\sin}(x)}{x}\)
  2. \(\displaystyle \ds \lim_{x\to 0} \dfrac{\ln (1+x)}{\cos x + e^x - 1}\)
  3. \(\displaystyle \ds \lim_{x\to -\infty} x\ln\lrpar{1 - \dfrac{1}{x}}\)
  4. \(\displaystyle \ds \lim_{x\to 0^+} (\sin 3x + 1)^{1/x}\)
Solution.
  1. \(\displaystyle \ds \lim_{x\to 0} \dfrac{\inv{\sin}(x)}{x} = 1\)
  2. \(\displaystyle \ds \lim_{x\to 0} \dfrac{\ln (1+x)}{\cos x + e^x - 1}=0\)
  3. \(\displaystyle \ds \lim_{x\to -\infty} x\ln\lrpar{1 - \dfrac{1}{x}}=-\dfrac{1}{2}\)
  4. \(\displaystyle \ds \lim_{x\to 0^+} (\sin 3x + 1)^{1/x} = e^{3\cot 1}\)
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