7.6 l’Hôpital’s Rule

7.6.1 Indeterminate forms 0=0 and \( \infty / \infty \)

A very useful tool to compute limits is l’Hôpita’s rule. However, as we shall see, there are instances where the rule cannot be immediately applied, or instances where the rule should not be applied from a logical point of view, or instances where the rule denitely yields no answer at all. Thus, when learning to apply l’Hôpita’s rule you must denitely be mindful of the context and how you apply it.

Theorem 1. Let \(f\) and \(g\) be differentiable functions on an open interval \(I\), and let \(c\in I\). Assume that
\[\lim_{x\to c}f(x) = 0\;\;\; \text{ and } \;\;\; \lim_{x\to c}g(x) = 0,\]
or, that
\[\lim_{x\to c}f(x) = \pm \infty\;\;\; \text{ and } \;\;\; \lim_{x\to c}g(x) =\pm \infty.\]
Assume further that \(g'(x) \neq 0\) for all \(x\in I\), except possibly at \(c\), and that
\[\lim_{x\to c}\frac{f'(x)}{g'(x)} =L, \;\; \text{ or } \;\; \lim_{x\to c}\frac{f'(x)}{g'(x)} =\infty, \;\; \text{ or } \;\; \lim_{x\to c}\frac{f'(x)}{g'(x)} =-\infty,\]
then, respectively,
\[\lim_{x\to c}\frac{f(x)}{g(x)} =L,\;\; \text{ or } \;\; \lim_{x\to c}\frac{f(x)}{g(x)} =\infty, \;\; \text{ or } \;\; \lim_{x\to c}\frac{f(x)}{g(x)} =-\infty,\]
that is
\[\lim_{x\to c}\frac{f(x)}{g(x)} =\lim_{x\to c}\frac{f'(x)}{g'(x)}.\]

Remark 1. The Theorem continues to be valid if the interval is of the form \((c, b)\) (or \((a, c)\)), and the limits are replaced with \(x\to c^+\) (respectively \(x\to c^{-}\)).

Example 1. Verify that l’Hôpital’s rule can be applied to evaluate the limit (a) \(\displaystyle \lim_{x\to 0}\frac{\ln \cos 2x}{\ln \cos 3x}\); (b) \(\displaystyle \lim_{x\to 0^+}\frac{\ln x}{1/x}\), and find the value of the limit.

Solution. 

(a) \(f(x) =\ln \cos 2x\) and \(g(x) = \ln \cos 3x\) are differentiable on \(\displaystyle I = \left (-\frac{\pi}{6}, \frac{\pi}{6}\right ) \) (explain why), and
\[\lim_{x\to 0}f(x) = \lim_{x\to 0}\ln \cos 2x = \ln 1 = 0\;\;\; \text{ and } \;\;\; \lim_{x\to 0}g(x) =\lim_{x\to 0}\ln \cos 3x = \ln 1 = 0.\]
Furthermore
\[g'(x) = -3\frac{\sin 3x}{\cos 3x}\neq 0 \;\, \forall \; x\in I-\{0\},\]
and
\[\lim_{x\to 0}\frac{f'(x)}{g'(x)} = \lim_{x\to 0}\frac{\displaystyle -2\frac{\sin 2x}{\cos 2x}}{\displaystyle -3\frac{\sin 3x}{\cos 3x}}
=\lim_{x\to 0} \frac{ \sin 2x}{\sin 3x}\cdot \frac{ 2\cos 3x}{ 3\cos 2x}
=\lim_{x\to 0} \frac{\displaystyle \frac{2\sin 2x}{2x}}{\displaystyle \frac{3\sin 3x}{3x}}\cdot \frac{ 2\cos 3x}{ 3\cos 2x}\underset{(*)}{ =} \frac{2}{3}\cdot\frac{2}{3}= \frac{4}{9}.\]
\((*)\) Here, we have used the trigonometric limit \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x} = 1\).

 
(b) Let \(f(x) =\ln x\) and \(\displaystyle g(x) = \frac{1}{x}\), then \(f\) and \(g\) are differentiable on the interval \(\displaystyle I = \left (0, \infty\right ) \), and
\[\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\ln x= -\infty\;\;\; \text{ and } \;\;\; \lim_{x\to 0^+}g(x) =\lim_{x\to 0^+}\frac{1}{x} = \infty.\]
Furthermore
\[g'(x) =- \frac{1}{x^2}\neq 0 \;\, \forall \; x\in I,\]
and
\[\lim_{x\to 0^+}\frac{f'(x)}{g'(x)} = \lim_{x\to 0^+}\frac{\displaystyle \frac{1}{x}}{\displaystyle -\frac{1}{x^2}}
=\lim_{x\to 0} -x
=0.\]
It follows that \(f\) and \(g\) satisfy the assumptions of Theorem 1, thus
\[\lim_{x\to 0^+}\frac{\ln x}{1/x} = \lim_{x\to 0}\frac{f'(x)}{g'(x)} = 0.\]
 

Solution. (a) \(f(x) =e^{1/x} – 1\) and \(\displaystyle g(x) = \frac{1}{x}\) are differentiable on the interval \(\displaystyle I = \left (0,\infty\right ) \), and
\[\lim_{x\to \infty }f(x) = \lim_{x\to \infty }e^{1/x} – 1 = e^0-1= 0\;\;\; \text{ and } \;\;\; \lim_{x\to \infty }g(x) =\lim_{x\to 0}\frac{1}{x} = 0.\]
In addition
\[g'(x) = -\frac{1}{x^2}\neq 0 \;\, \forall \; x\in I,\]
and
\[\lim_{x\to \infty }\frac{f'(x)}{g'(x)} = \lim_{x\to \infty }\frac{\displaystyle -e^{1/x}\cdot \frac{1}{x^2} }{\displaystyle -\frac{1}{x^2}} =\lim_{x\to \infty } e^{1/x} =e^0= 1.\]
It follows that \(f\) and \(g\) satisfy the assumptions of Theorem 2, thus
\[\lim_{x\to \infty }\frac{e^{1/x}-1}{1/x} = \lim_{x\to \infty }\frac{f'(x)}{g'(x)} = 1.\]

(b) Let \(f(x) =x^2\) and \(\displaystyle g(x) =e^x\), then \(f\) and \(g\) are differentiable on \(\displaystyle \mathbb{R}\), and
\[\lim_{x\to \infty }f(x) = \lim_{x\to \infty }x^2= \infty\;\;\; \text{ and } \;\;\; \lim_{x\to \infty }g(x) =\lim_{x\to \infty}e^x = \infty.\]
In addition
\[g'(x) =e^x\neq 0 \;\, \forall \; x\in \mathbb{R},\]
and
\[\lim_{x\to \infty }\frac{f'(x)}{g'(x)} = \lim_{x\to \infty }\frac{2x}{e^x}.\]
Note that the last limit leads to the indeterminate form \(\displaystyle \frac{\infty}{\infty}\). Thus, we can apply l’Hôpital’s rule to this limit (verify that this is so).
\[\lim_{x\to \infty }\frac{f'(x)}{g'(x)} = \lim_{x\to \infty }\frac{2x}{e^x} =\lim_{x\to \infty }\frac{f”(x)}{g”(x)} = \lim_{x\to \infty }\frac{2}{e^x} =0.\]
Hence
\[\lim_{x\to \infty }\frac{x^2}{e^x} = \lim_{x\to \infty }\frac{f'(x)}{g'(x)} = \lim_{x\to \infty }\frac{2x}{e^x} =\lim_{x\to \infty }\frac{f”(x)}{g”(x)} = \lim_{x\to \infty }\frac{2}{e^x} =0.\]

7.6.2 Indeterminate form \(0 \cdot \infty\)

As pointed out in Remark 2 and Remark 3, it is not always suitable to apply l’Hôpita’s rule directly. In this and the next part, we consider contexts where in order to apply l’Hôpita’s rule, it is necessary to analyze the limit and find a way to rewrite it to be able to use the rule. Here we consider the indeterminate form \(0\cdot \infty\). The way to deal with this indeterminate form is by rewriting it into an equivalent form \(\displaystyle \frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\). The algebraic principle behind this is simple:
\[0\cdot \infty \;\; \longleftrightarrow \;\; \frac{ \infty}{\displaystyle \frac{1}{0}}\;\; \longleftrightarrow \;\; \frac{ \infty}{\displaystyle \infty}
\hspace{.3in} \text{ or } \hspace{.3in}
0\cdot \infty \;\; \longleftrightarrow \;\; \frac{ 0}{\displaystyle \frac{1}{\infty}}\;\; \longleftrightarrow \;\; \frac{ 0}{\displaystyle 0}.
\]
Thus, you will have to be mindful when deciding how to rewrite the indeterminate form.

Example 1.  Compute the limit.    (a) \(\displaystyle \lim_{x\to 0^+ } x^3\ln x\);    (b) \(\displaystyle \lim_{x\to \infty } xe^{-x}\);    (c) \(\displaystyle \lim_{x\to 0^+ } \sin x \log_2 x\).

Solution. (a) Note that the limit takes the indeterminate form \(0\cdot \infty\), and that can be rewritten as
\[\lim_{x\to 0^+ } x^3\ln x = \lim_{x\to 0^+ } \frac{\ln x}{x^{-3}},\]
and, that it thus takes the indeterminate form \(\displaystyle -\frac{\infty}{\infty}\). Furthermore, it can be verify that l’Hôpital’s rule can be applied (verify this). Hence
\[\lim_{x\to 0^+ } x^3\ln x = \lim_{x\to 0^+ } \frac{\ln x}{x^{-3}} =\lim_{x\to 0^+ } \frac{x^{-1}}{-3x^{-4}} = \lim_{x\to 0^+ } \frac{2x^3}{-3} = 0. \]

 
(b) Given that \(\displaystyle \lim_{x\to \infty } e^{-x}=0\), the limit corresponds to the indeterminate form \(\infty\cdot 0\). It can be rewritten as follows:
\[ \lim_{x\to \infty } xe^{-x} = \lim_{x\to \infty } \frac{x}{e^x}\; \underset{\text{l’H}}{=} \;\lim_{x\to \infty } \frac{1}{e^x} = 0.\]
Note that the limit could have been rewritten in an alternative form, but not also that this new form would not have helped:
\[ \lim_{x\to \infty } x e^{-x} = \lim_{x\to \infty } \frac{e^{-x}}{\displaystyle \frac{1}{x}}\; \underset{\text{l’H}}{=} \;\lim_{x\to \infty } \frac{-e^{-x}}{\displaystyle -\frac{1}{x^2}}\;\; (!).\]

 
(c) Since \(\sin0=0\), \(\displaystyle \lim_{x\to 0^+ } \log_2 x = -\infty\), and \(\displaystyle \log_2x= \frac{\ln x}{\ln 2}\), we have
\[\lim_{x\to 0^+ } \sin x \log_2 x = \lim_{x\to 0^+ } \frac{ \ln x}{\ln 2 \csc x} \; \underset{\text{l’H}}{=} \; \lim_{x\to 0^+ } \frac{x^{-1}}{-\ln 2 \csc x \cot x} = \lim_{x\to 0^+ } \frac{\sin x}{x}\cdot \frac{\tan x}{-\ln 2} = 1\cdot 0 = 0.\]

7.6.4 Indeterminate forms \( 0^{0}\), \(\infty^{0}\), and \(1^{\infty}\)

The way to deal with the indeterminate forms \(0^0\), \(0^\infty\), \(\infty^0\), and \(1^\infty\) is by rewriting them –by using natural logarithms– into an equivalent form \(0\cdot\infty\), and then rewrite again into either one of the forms \(\displaystyle \frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\).

Example 1. Evaluate the limit.    (a) \(\displaystyle \lim_{x\to 0^+ } x^{x}\);    (b) \(\displaystyle \lim_{x\to \infty } x^{1/x}\);    (c) \(\displaystyle \lim_{x\to \infty } \left ( 1 + \frac{1}{x}\right )^x\).

Solution. (a) Clearly, this limit corresponds to the indeterminate form \(0^0\). Thus, we rewrite as follows:
\[x^x = e^{\ln x^x} = e^{x\ln x}. \;\; \Longrightarrow \;\; \lim_{x\to 0^+ } x^{x} = \lim_{x\to 0^+ } e^{x\ln x} \;\; \underset{(*)}{=}\;\; \exp\left (\lim_{x\to 0^+ } x\ln x\right )\]
\((^*)\) by continuity of the exponential function \(y=e^x\). And, we have
\[\lim_{x\to 0^+ } x\ln x = \lim_{x\to 0^+ } \frac{\;\;\ln x\;\;}{\displaystyle \frac{1}{x}} =
\;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to 0^+ } \frac{\displaystyle \frac{1}{x}}{\displaystyle -\frac{1}{x^2}}
= \lim_{x\to 0^+ } -x = 0. \]
Thus
\[ \lim_{x\to 0^+ } x^{x} =\exp\left (\lim_{x\to 0^+ } x\ln x\right ) = e^0=1.\]

\noindent
(b) Since \(\displaystyle \lim_{x\to \infty } \frac{1}{x} = 0\), this limit correspond to the indeterminate form \(\infty^0\). Again, we have, by continuity of the exponential function,
\[x^{1/x} = e^{\ln x^{1/x}} = e^{(\ln x)/x}. \;\;
\Longrightarrow \;\; \lim_{x\to \infty } x^{1/x} = \lim_{x\to \infty } e^{(\ln x)/x} = \exp\left (\lim_{x\to\infty } \frac{\ln x}{x}\right ) .\]
And
\[ \lim_{x\to\infty } \frac{\ln x}{x}
\;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to\infty } \frac{\displaystyle \;\;\frac{1}{x}\;\;}{1} = 0.\;\; \Longrightarrow \;\; \lim_{x\to \infty } x^{1/x} =\exp\left (\lim_{x\to\infty } \frac{\ln x}{x}\right ) = e^0 = 1.
\]

\noindent
(c) In this case, the limit corresponds to the indeterminate form \(1^{\infty}\).
\[\left(1+ \frac{1}{x}\right )^x = \exp\left [\ln \left(1+ \frac{1}{x}\right )^x\,\right ] =
\exp\left [x \ln \left(1+ \frac{1}{x}\right )\right ].\]
Thus
\[\lim_{x\to \infty } \left(1+ \frac{1}{x}\right )^x =\exp\left [\lim_{x\to\infty } x \ln \left(1+ \frac{1}{x}\right ) \right] .\]
Now then
\[
\lim_{x\to\infty } x \ln \left(1+ \frac{1}{x}\right )
\;\;\underset{(\infty\cdot 0)}{=}\;\;
\lim_{x\to\infty } \frac{\displaystyle \ln \left(1+ \frac{1}{x}\right ) }{\displaystyle \frac{1}{x}}
\;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to\infty } \frac{\displaystyle \frac{1}{1+\frac{1}{x}} \cdot \left ( -\frac{1}{x^2}\right ) }{\displaystyle -\frac{1}{x^2}} = 1.\;\; \Longrightarrow\]
\[\Longrightarrow \;\;
\lim_{x\to \infty } \left(1+ \frac{1}{x}\right )^x =\exp\left [\lim_{x\to\infty } x \ln \left(1+ \frac{1}{x}\right ) \right] = e^1 = e.\]

7.6.5 Indeterminate form \(\infty – \infty\)

 
The way to deal with an indeterminate form \(\infty – \infty\) is either by rewriting it into an equivalent form \(\displaystyle \frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\), or by rewriting into a form that can be analyzed to render the limit.

Example 1.  Compute the limit. Compute the limit.    (a) \(\displaystyle \lim_{x\to 0^+ } \left (\frac{1}{x}-\frac{1}{\tan x}\right )\);    (b) \(\displaystyle \lim_{x\to \infty }( x – \ln x)\);    (c) \(\displaystyle \lim_{x\to \infty }\left ( x^2- x^3\right )\),    (d) \(\displaystyle \lim_{x\to 0^+ }\left ( \csc x + \log_2 x^2\right )\).

(a)
\begin{align*}
\lim_{x\to 0^+ } \left (\frac{1}{x}-\frac{1}{\tan x}\right )
& \;\;\underset{\hphantom{\text{l’H}}}{=}\;\;
\lim_{x\to 0^+ } \left (\frac{1}{x}-\frac{\cos x}{\sin x}\right ) =
\lim_{x\to 0^+ } \frac{\sin x- x\cos x}{x\sin x}
\vphantom{\frac{A}{\frac{A}{A}}}= \\ & \;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to 0^+ } \frac{\cos x- \cos x + x \sin x}{\sin x + x\cos x}
=
\lim_{x\to 0^+ } \frac{ x \sin x}{\sin x + x\cos x}
\vphantom{\frac{A}{\frac{A}{A}}}= \\ & \;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to 0^+ } \frac{ \sin x + x\cos x}{\cos x + \cos x – x\sin x} = \frac{0}{2} =0.
\end{align*}
(b) Using the fact that \(\displaystyle \lim_{x\to\infty } \frac{\ln x}{x} = 0\) from Example 1 in 7.6.5, we have
\[\lim_{x\to \infty }( x – \ln x)
=
\lim_{x\to \infty }x\left ( 1 – \frac{\ln x}{x}\right ) = \infty(1-0)= \infty.\]
(c)
\[\lim_{x\to \infty }\left ( x^2- x^3\right ) = \lim_{x\to \infty }x^3\left ( \frac{1}{x}-1\right ) = -\infty.\]
(d)
\[
\lim_{x\to 0^+ }\left ( \csc x + \log_2 x^2\right )
=
\lim_{x\to 0^+ }\left ( \frac{1}{\sin x} + \frac{2\ln x}{\ln 2}\right ) =
\lim_{x\to 0^+ } \frac{\ln 2 + 2\sin x\ln x}{\ln 2 \sin x}.
\]
To analyze this limit, we evaluate
\[
\lim_{x\to 0^+ } \sin x\ln x
= \lim_{x\to 0^+ } \frac{\ln x}{\csc x}
\;\;\underset{\text{l’H}}{=}\;\;
\lim_{x\to 0^+ } \frac{\displaystyle \frac{1}{x}}{-\csc x\cot x}
= \lim_{x\to 0^+ } -\frac{\sin x}{x}\tan x = 0,
\]
\[\Longrightarrow \;\; \lim_{x\to 0^+ }\big(\ln 2 + 2\sin x\ln x\big) = \ln 2.\]
And
\[\lim_{x\to 0^+ } \ln 2 \sin x = 0.\]
In addition, since \(\ln 2 < 0\), both, numerator and denominator are negative, thus \[\lim_{x\to 0^+ }\left ( \csc x + \log_2 x^2\right ) = \lim_{x\to 0^+ } \frac{\ln 2 + 2\sin x\ln x}{\ln 2 \sin x} = \infty.\]