\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \LARGE L'H\^{o}pital's Rule} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2010} \rfoot{\small Last revised: March 16, 2014} \textwidth 7.5in \textheight 9.7in \setlength{\headheight}{25pt} \setlength{\parindent}{0in} \begin{document} Suppose that $f$ and $g$ are differentiable and $g^{\prime }\left( x\right) \not=0$ on an open interval that contains $a$ (except possibly at $a$). \ Suppose that \begin{equation*} \lim\limits_{x\rightarrow a}f\left( x\right) =0\text{ and }% \lim\limits_{x\rightarrow a}g\left( x\right) =0 \end{equation*}% or that% \begin{equation*} \lim\limits_{x\rightarrow a}f\left( x\right) =\pm \infty \text{ and }% \lim\limits_{x\rightarrow a}g\left( x\right) =\pm \infty \end{equation*}% (In other words, we have an indeterminate form of type $\dfrac{0}{0}$ or $% \pm \dfrac{\infty }{\infty }$. \ Then \begin{equation*} \lim\limits_{x\rightarrow a}\dfrac{f\left( x\right) }{g\left( x\right) }% =\lim\limits_{x\rightarrow a}\dfrac{f^{\prime }\left( x\right) }{g^{\prime }\left( x\right) } \end{equation*}% If the limit on the right side exists. \bigskip \bigskip \begin{center} {\Large Sample Problems\bigskip } \end{center} %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x^{2}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x+\sin x}$ \item $\lim\limits_{x\rightarrow \pi ^{-}}\dfrac{\sin x}{1-\cos x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x+2\sin x}$ \item $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\ln x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{2\sin x-\sin 2x}{x-\sin x}$ \item $\lim\limits_{x\rightarrow 0^{+}}x\ln x$ \item $\lim\limits_{x\rightarrow 0}\dfrac{5^{x}-1}{x^{3}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x^{2}}$ \item $\lim\limits_{x\rightarrow 1}\dfrac{\ln x}{x^{2}-x}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{4x-\sin 4x}{x^{3}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin x-\tan x}{x^{3}}$ \item $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}-\ln x}{\sqrt[3]{x}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{e^{x^{2}}+10}{1-\cos x}$ \ \end{enumerate} %TCIMACRO{\TeXButton{2 col end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \bigskip \begin{center} {\Large Practice Problems\bigskip } \end{center} %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\lim\limits_{x\rightarrow 0^{+}}\dfrac{e^{x}-1}{x^{2}}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{e^{3x}-1}{5x}$ \item $\lim\limits_{a\rightarrow 1}\dfrac{3a^{2}-2a-1}{5a^{2}-a-4}$ \item $\lim\limits_{y\rightarrow \infty }\dfrac{\ln y}{\sqrt[3]{y}}$ \item $\lim\limits_{x\rightarrow \pi }\dfrac{\sin x}{x-\pi }$ \item $\lim\limits_{m\rightarrow 2}\dfrac{m^{5}-32}{m^{3}-8}$ \item $\lim\limits_{\theta \rightarrow \pi /2}\dfrac{\tan \theta }{\tan 5\theta }$ \item $\lim\limits_{x\rightarrow \infty }\dfrac{x}{\ln \left( x+1\right) }$ \item $\lim\limits_{x\rightarrow 0}\dfrac{x^{3}}{\tan x-x}$ \item $\lim\limits_{\beta \rightarrow 0}\dfrac{\sin \beta -\beta }{\tan \beta -\beta }$ \item $\lim\limits_{x\rightarrow 0}\left( x^{2}e^{1/x^{2}}\right) $ \item $\lim\limits_{p\rightarrow 0}\dfrac{e^{3p}-1}{\sin 2p}$ \item $\lim\limits_{x\rightarrow 0}\dfrac{e^{\left( x^{2}\right) }+10}{% 1-\cos x}$ \item $\lim\limits_{x\rightarrow 1}\dfrac{x^{2/3}-x^{1/2}}{x-1}$ \item $\lim\limits_{\alpha \rightarrow 0}\dfrac{\alpha }{\arctan 2\alpha }$ \ \end{enumerate} %TCIMACRO{\TeXButton{2 col end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \begin{center} {\Large Sample Problems - Answers\bigskip } \end{center} 1.) \ $\infty $ \ \ \ \ 2.) \ $0$ \ \ \ \ 3.) \ $0$\ \ \ \ \ 4.) \ $\dfrac{1% }{3}$ \ \ \ \ \ 5.) \ $\infty $ \ \ \ \ \ 6.) \ $6$ \ \ \ \ \ 7.) \ $0$ \ \ \ \ \ 8.) \ $\infty $ \ \ \ \ \ \ 9.) \ $-\dfrac{1}{2}$ \ \ \ \ \ \ \ 10.) \ $1$ \medskip 11.) \ $\dfrac{32}{3}$ \ \ \ \ \ \ \ 12.) \ $-\dfrac{1}{2}$ \ \ \ \ \ 13.) \ $\infty $ \ \ \ \ \ 14.) \ $\infty $\bigskip \bigskip \begin{center} {\Large Practice Problems - Answers\bigskip }\bigskip \end{center} 1.) $\ \infty $ \ \ \ 2.) $\ \dfrac{3}{5}$ \ \ \ 3.) $\ \dfrac{4}{9}$ \ \ \ 4.) $\ 0$ \ \ \ 5.) $\ -1$ \ \ \ \ 6.) $\ \dfrac{20}{3}$ \ \ \ 7.) $\ 5$ \ \ \ 8.) $\ \infty $ \ \ \ \ 9.) $\ 3$ \ \ \ \ 10.) $\ -\dfrac{1}{2}$ \ \ 11.) $% \ \infty $\medskip 12.) $\ \dfrac{3}{2}$ \ \ \ \ 13.) $\ \infty $ \ \ \ This is NOT an indeterminate! \ \ \ \ 14.) $\dfrac{1}{6}$ \ \ \ \ 15.) $\ \dfrac{1}{2}$% \bigskip \bigskip \begin{center} {\Large Sample Problems - Solutions\bigskip }\bigskip \end{center} \begin{enumerate} \item $\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x^{2}}$ Solution: \ This is an $\dfrac{\infty }{\infty }$ type of an indeterminate, so we can apply l'H\^{o}pital's rule. \ We differentiate both numerator and denominator:% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x^{2}}=\lim\limits_{x% \rightarrow \infty }\dfrac{e^{x}}{2x} \end{equation*}% $\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x^{2}}$ is an $\dfrac{% \infty }{\infty }$ type of an indeterminate, so we can apply l'H\^{o}pital's rule again% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{2x}=\lim\limits_{x% \rightarrow \infty }\dfrac{e^{x}}{2}=\fbox{$\infty $} \end{equation*}% $\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{2}$ is no longer an indeterminate because the numerator approaches infinity while the denominator approaches $2.$ \ This limit is $\infty $.\medskip \item $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x+\sin x}$ Solution: \ $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x+\sin x}$ is a $% \dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x+\sin x}=\lim\limits_{x% \rightarrow 0}\dfrac{-\sin x}{1+\cos x}=\dfrac{0}{2}=\fbox{$0$} \end{equation*} \item $\lim\limits_{x\rightarrow \pi ^{-}}\dfrac{\sin x}{1-\cos x}$ Solution: \ This limit does not qualify for l'H\^{o}pital's rule because substituting $\pi $ into the expression does NOT result in an indeterminate. \ \begin{equation*} \lim\limits_{x\rightarrow \pi ^{-}}\dfrac{\sin x}{1-\cos x}=\dfrac{0}{% 1-\left( -1\right) }=\dfrac{0}{2}=\fbox{$0$} \end{equation*} If applied the rule, we would get an incorrect answer, $-\infty $. \ \textbf{% So, it is very important to check for the conditions of the rule.}\medskip \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x+2\sin x}$ Solution: \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x+2\sin x}$ a $% \dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x+2\sin x}=\lim\limits_{x% \rightarrow 0}\dfrac{\cos x}{1+2\cos x} \end{equation*}% $\lim\limits_{x\rightarrow 0}\dfrac{\cos x}{1+2\cos x}$ is no longer an indeterminate: we get $\dfrac{1}{3}$ when we substitute $x=0$.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x}{1+2\cos x}=\dfrac{1}{1+2}=\fbox{$% \dfrac{1}{3}$} \end{equation*}% Note that we didn't really need L'H\^{o}pital's rule to compute this limit. \ Recall that $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$ and then of course its reciproal approaches $1$ as well: $\lim\limits_{x\rightarrow 0}% \dfrac{x}{\sin x}=1$. \ We can compute this limit by dividing both numerator and denominator by $\sin x$.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x+2\sin x}=\lim\limits_{x% \rightarrow 0}\dfrac{1}{\dfrac{x}{\sin x}+2}=\dfrac{1}{\lim\limits_{x% \rightarrow 0}\dfrac{x}{\sin x}+2}=\dfrac{1}{1+2}=\dfrac{1}{3} \end{equation*} \item $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\ln x}$ Solution: \ $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\ln x}$ is an $\dfrac{\infty }{\infty }$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\ln x}=\lim\limits_{x% \rightarrow \infty }\dfrac{\dfrac{1}{2\sqrt{x}}}{\dfrac{1}{x}}% =\lim\limits_{x\rightarrow \infty }\dfrac{1}{2\sqrt{x}}\dfrac{x}{1}% =\lim\limits_{x\rightarrow \infty }\dfrac{1}{2}\sqrt{x}=\fbox{$\infty $} \end{equation*} \item $\lim\limits_{x\rightarrow 0}\dfrac{2\sin x-\sin 2x}{x-\sin x}$ Solution: \ We substitute $x$ into the expression and get a $\dfrac{0}{0}$ type of an indeterminate. \ So we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{2\sin x-\sin 2x}{x-\sin x}% =\lim\limits_{x\rightarrow 0}\dfrac{2\cos x-\cos 2x\left( 2\right) }{1-\cos x% }=\lim\limits_{x\rightarrow 0}\dfrac{2\left( \cos x-\cos 2x\right) }{1-\cos x% } \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{2\left( \cos x-\cos 2x\right) }{1-\cos x}% =\lim\limits_{x\rightarrow 0}\dfrac{2\left( -\sin x+\sin 2x\left( 2\right) \right) }{\sin x}=\lim\limits_{x\rightarrow 0}\dfrac{2\left( -\sin x+2\sin 2x\right) }{\sin x} \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{2\left( -\sin x+2\sin 2x\right) }{\sin x}% =\lim\limits_{x\rightarrow 0}\dfrac{2\left( -\cos x+2\cos 2x\left( 2\right) \right) }{\cos x}=\lim\limits_{x\rightarrow 0}\dfrac{2\left( -\cos x+4\cos 2x\right) }{\cos x} \end{equation*}% This is no longer an indeterminate:% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{2\left( -\cos x+4\cos 2x\right) }{\cos x}=% \dfrac{2\left( -1+4\right) }{1}=\fbox{$6$} \end{equation*} \item $\lim\limits_{x\rightarrow 0^{+}}x\ln x$ Solution: \ Although it doesn't appear so, this is either a $\dfrac{0}{0}$ or an $\dfrac{\infty }{\infty }$ type of an indeterminate after we re-write it: \ $\lim\limits_{x\rightarrow 0^{+}}\dfrac{~\ln x~}{~\dfrac{1}{x}~}$ is an $\dfrac{-\infty }{\infty }$ type of an indeterminate\ and $% \lim\limits_{x\rightarrow 0^{+}}\dfrac{~x~}{~\dfrac{1}{\ln x}~}$ is a $% \dfrac{0}{0}$ type of an indeterminate. \ We will use the first form because it seems to yield for slightly easier computation.% \begin{equation*} \lim\limits_{x\rightarrow 0^{+}}x\ln x=\lim\limits_{x\rightarrow 0^{+}}% \dfrac{~\ln x~}{~\dfrac{1}{x}~}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{~% \dfrac{1}{x}~}{~-\dfrac{1}{x^{2}}~}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{1% }{x}\left( -\dfrac{x^{2}}{1}\right) =\lim\limits_{x\rightarrow 0^{+}}\left( -x\right) =\fbox{$0$} \end{equation*} \item $\lim\limits_{x\rightarrow 0}\dfrac{5^{x}-1}{x^{3}}$ Solution: \ This is a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{5^{x}-1}{x^{3}}=\lim\limits_{x\rightarrow 0}\dfrac{\left( \ln 5\right) 5^{x}}{3x^{2}}=\fbox{$\infty $} \end{equation*}% In $\lim\limits_{x\rightarrow 0}\dfrac{\left( \ln 5\right) 5^{x}}{3x^{2}}$ the numerator approaches $1$ while the denominator approaches zero and is positive. \ This limit is $\infty $.\medskip \item $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x^{2}}$ Solution: \ \ This is a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x^{2}}=\lim\limits_{x% \rightarrow 0}\dfrac{-\sin x}{2x} \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{-\sin x}{2x}=\lim\limits_{x\rightarrow 0}% \dfrac{-\cos x}{2}=\fbox{$-\dfrac{1}{2}$} \end{equation*} \item $\lim\limits_{x\rightarrow 1}\dfrac{\ln x}{x^{2}-x}$ Solution: \ \ This is a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 1}\dfrac{\ln x}{x^{2}-x}=\lim\limits_{x\rightarrow 1}\dfrac{\dfrac{1}{x}}{2x-1}=\dfrac{1}{2-1}=\fbox{$1$} \end{equation*}% \pagebreak \item $\lim\limits_{x\rightarrow 0}\dfrac{4x-\sin 4x}{x^{3}}$ Solution: \ \ This is a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{4x-\sin 4x}{x^{3}}=\lim\limits_{x% \rightarrow 0}\dfrac{4-4\cos 4x}{3x^{2}} \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{4-4\cos 4x}{3x^{2}}=\lim\limits_{x% \rightarrow 0}\dfrac{16\sin 4x}{6x}=\lim\limits_{x\rightarrow 0}\dfrac{8\sin 4x}{3x} \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{8\sin 4x}{3x}=\lim\limits_{x\rightarrow 0}% \dfrac{32\cos 4x}{3}=\fbox{$\dfrac{32}{3}$} \end{equation*} \item $\lim\limits_{x\rightarrow 0}\dfrac{\sin x-\tan x}{x^{3}}=-\dfrac{1}{2} $ Solution: \ \ This is a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H\^{o}pital's rule. \ \newline Recall that $\dfrac{d}{dx}\tan x=\dfrac{1}{\cos ^{2}x}$% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{\sin x-\tan x}{x^{3}}=\lim\limits_{x% \rightarrow 0}\dfrac{\cos x-\dfrac{1}{\cos ^{2}x}}{3x^{2}}% =\lim\limits_{x\rightarrow 0}\dfrac{\cos x-\left( \cos x\right) ^{-2}}{3x^{2}% } \end{equation*}% This is still a $\dfrac{0}{0}$ type of an indeterminate, so we can apply l'H% \^{o}pital's rule again.% \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x-\left( \cos x\right) ^{-2}}{3x^{2}} &=&\lim\limits_{x\rightarrow 0}\dfrac{-\sin x-\left( -2\right) \left( \cos x\right) ^{-3}\left( -\sin x\right) }{6x}=\lim\limits_{x\rightarrow 0}\dfrac{% -\sin x-2\sin x\left( \cos x\right) ^{-3}}{6x} \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\sin x\left( -1-\dfrac{2}{\cos ^{3}x}% \right) }{6x}=\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}% \lim\limits_{x\rightarrow 0}\dfrac{\left( -1-\dfrac{2}{\cos ^{3}x}\right) }{6% }=1\cdot \dfrac{-1-2}{6}=\fbox{$-\dfrac{1}{2}$} \end{eqnarray*}% If we didn't notice the limit $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}$ in the expressions, we can still get the answer by applying l'H\^{o}pital's rule one more time.% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{-\sin x-2\sin x\left( \cos x\right) ^{-3}% }{6x}% =~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{equation*}% \begin{eqnarray*} &=&\lim\limits_{x\rightarrow 0}\dfrac{\sin x\left( -1-2\left( \cos x\right) ^{-3}\right) }{6x}=\lim\limits_{x\rightarrow 0}\dfrac{\cos x\left( -1-2\left( \cos x\right) ^{-3}\right) +\sin x\left( -2\left( -3\right) \left( \cos x\right) ^{-4}\left( -\sin x\right) \right) }{6} \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{\cos x\left( -1-\dfrac{2}{\cos ^{3}x}% \right) +\sin x\left( -6\dfrac{\sin x}{\cos ^{4}x}\right) }{6}=\dfrac{1\cdot \left( -1-\dfrac{2}{1}\right) +0\cdot \left( -6\dfrac{0}{1}\right) }{6}=-% \dfrac{1}{2} \end{eqnarray*}% \pagebreak \item $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}-\ln x}{\sqrt[3]{x}}% =\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{2\sqrt{x}}-\dfrac{1}{x}}{% \dfrac{1}{3}x^{-2/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{\sqrt{x% }}{2}-1}{\dfrac{1}{3}x^{1/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{% \dfrac{1}{4\sqrt{x}}}{\dfrac{1}{9}x^{-2/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{\sqrt{x}}{4}}{\dfrac{1}{9}x^{1/3}}=\infty $ Solution: \ In case of this problem, it will take some work to determine whether we can use L'H\^{o}pital's rule for this limit. \ The numerator itself is an $\infty -\infty $ type of an indeterminate. \ We first factor out $\sqrt{x}$.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( \sqrt{x}-\ln x\right) =\lim\limits_{x\rightarrow \infty }\sqrt{x}\left( 1-\dfrac{\ln x}{\sqrt{x}}% \right) \end{equation*}% Inside the parentheses, $\dfrac{\ln x}{\sqrt{x}}$ is an $\dfrac{\infty }{% \infty }$ type of an indeterminate. \ We apply l'H\^{o}pital's rule:% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{\sqrt{x}}=\lim\limits_{x% \rightarrow \infty }\dfrac{~\dfrac{1}{x}~}{\dfrac{1}{2\sqrt{x}}}% =\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}\dfrac{2\sqrt{x}}{1}% =\lim\limits_{x\rightarrow \infty }\dfrac{2}{\sqrt{x}}=0 \end{equation*}% This the numerator is \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( \sqrt{x}-\ln x\right) =\lim\limits_{x\rightarrow \infty }\sqrt{x}\left( 1-\dfrac{\ln x}{\sqrt{x}}% \right) =\lim\limits_{x\rightarrow \infty }\sqrt{x}\lim\limits_{x\rightarrow \infty }\left( 1-\dfrac{\ln x}{\sqrt{x}}\right) =\left( \lim\limits_{x\rightarrow \infty }\sqrt{x}\right) \left( 1-0\right) =\infty \end{equation*}% Thus the numerator approaches $\infty ,$ and so we have an $\dfrac{\infty }{% \infty }$ type of an indeterminate. \ We apply l'H\^{o}pital's rule.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}-\ln x}{\sqrt[3]{x}}% =\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{2\sqrt{x}}-\dfrac{1}{x}}{% \dfrac{1}{3}x^{-2/3}} \end{equation*}% We multiply both numerator and denominator by $x:$% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{2\sqrt{x}}-\dfrac{1}{x}}{% \dfrac{1}{3}x^{-2/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{\sqrt{x% }}{2}-1}{\dfrac{1}{3}x^{1/3}} \end{equation*}% This is still an $\dfrac{\infty }{\infty }$ type of an indeterminate, so we apply l'H\^{o}pital's rule again.% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{\sqrt{x}}{2}-1}{\dfrac{1}{3}% x^{1/3}} &=&\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{4\sqrt{x}}}{% \dfrac{1}{9}x^{-2/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{4}% x^{-1/2}}{\dfrac{1}{9}x^{-2/3}}=\lim\limits_{x\rightarrow \infty }\dfrac{% 9x^{-1/2-\left( -2/3\right) }}{4}=\lim\limits_{x\rightarrow \infty }\dfrac{9% }{4}x^{\tfrac{2}{3}-\tfrac{1}{2}}=\lim\limits_{x\rightarrow \infty }\dfrac{9% }{4}x^{\tfrac{1}{6}} \\ &=&\lim\limits_{x\rightarrow \infty }\dfrac{9}{4}\sqrt[6]{x}=\fbox{$\infty $} \end{eqnarray*} \item $\lim\limits_{x\rightarrow 0}\dfrac{e^{x^{2}}+10}{1-\cos x}$ This limit can not be solved using l'H\^{o}pital's rule because it is not an indeterminate. \ The numerator approaches $11$, the denominator approaches $% 0^{+}$ and so the quotient approaches $\infty $. \ If we apply l'H\^{o}% pital's rule, we will get a wrong result:% \begin{equation*} \lim\limits_{x\rightarrow 0}\dfrac{e^{x^{2}}+10}{1-\cos x}% =\lim\limits_{x\rightarrow 0}\dfrac{2xe^{x^{2}}}{\sin x}=\lim\limits_{x% \rightarrow 0}\dfrac{x}{\sin x}\cdot \lim\limits_{x\rightarrow 0}2e^{x^{2}}=1\cdot 2=2\text{ \ \ \ incorrect} \end{equation*}% Before applying l'H\^{o}pital's rule, always check first whether the conditions for it hold. \end{enumerate} \bigskip \bigskip \bigskip {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}