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%  L'Hopital's Rule                                %
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\problemgrouptitle{L'H\^{o}pital's Rule}

\begin{problemblock}[2]{Evaluate the given limit if it exists.  If it does not exist but has an infinite trend ($+\infty$ or $-\infty$) indicate the trend.}

\addtoanswerkey{\begin{multicols}{4}}
\item $\displaystyle\lim_{ x \to 2}{ \frac{2x^2-x-6}{x^2+x-6}}$
\answer{
$\displaystyle\frac 75$
}

\item $\displaystyle\lim_{ x \to 0}{ \frac{\sin 3x}{5x}}$
\answer{
$\displaystyle\frac 35$
}

\item $\displaystyle\lim_{ x \to \frac{\pi}{2}}{ \frac{1-\sin x}{2 \cos x}}$
\answer{
$\displaystyle 0$
}

\item $\displaystyle\lim_{ x \to 0}{ \frac{\tan^{-1} x}{5x}}$
\answer{
$\displaystyle\frac 15$
}

\item $\displaystyle\lim_{ x \to \infty}{ \frac{\ln x}{x^2}}$
\answer{
$\displaystyle 0$
}

\item $\displaystyle\lim_{ x \to \infty}{ \frac{3e^x+\ln x}{e^x+x}}$
\answer{
$\displaystyle 3$
}

\item $\displaystyle\lim_{ x \to \infty}{(x-2) e^{-x^3}}$
\answer{
$\displaystyle 0$
}

\item $\displaystyle\lim_{ x \to \infty}{ e^{-2x} \ln x}$
\answer{
$\displaystyle 0$
}

\item $\displaystyle\lim_{ x \to 0}{\left( \frac 2{e^x-1}-\frac 3x\right)}$
\answer{
$\displaystyle -\infty$
}

\item $\displaystyle\lim_{ x \to -\infty}{x \left(\frac {\pi}2+\tan^{-1} x \right)}$
\answer{
$\displaystyle -1$
}

\item $\displaystyle\lim_{ x \to 1}{\left( \frac 1{x-1}-\frac 1{x^2-1}\right)}$
\answer{
$\displaystyle \infty$
}

\item $\displaystyle\lim_{ x \to \infty}{ \left[\ln x -\ln (2x+3)\right]}$
\answer{
$\displaystyle \ln \frac 12$
}

\item $\displaystyle\lim_{ x \to \infty}{\left( 1-\frac 5x \right)^{2x}}$
\answer{
$\displaystyle e^{-10}$
}

\item $\displaystyle\lim_{ x \to 0}{ \left( e^{2x}-1\right)^{3x}}$
\answer{
$\displaystyle 1$
}

\item $\displaystyle\lim_{ x \to \infty}{ \left(1+\ln x \right)^{e^{-x}}}$
\answer{
$\displaystyle 1$
}

\item $\displaystyle\lim_{ x \to 0}{\left ( \sin 2x \right)^x}$
\answer{
$\displaystyle 1$
}

\addtoanswerkey{\end{multicols}}
\end{problemblock}

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