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\problemgrouptitle{Limits at Infinity and Horizontal Asymptotes}

\begin{problemblock}[2]{Determine the following limits.  For any limit that does not exist, identify if it has an infinite trend ($\infty$ or $-\infty$).}

\addtoanswerkey{\begin{multicols}{2}}

\item $\displaystyle\lim_{x\to-\infty}\displaystyle\frac{18x^2-3x}{3x^5-3x^2+2}$
\answer{$0$}

\item $\displaystyle\lim_{x\to\infty}\displaystyle\frac{x^2}{\displaystyle\sqrt{x^3+2}}$
\answer{$\infty$}

\item $\displaystyle\lim_{x\to-\infty}\displaystyle\frac{\displaystyle\sqrt{x^2+2x}}{6x+3}$
\answer{$\displaystyle-\frac{1}{6}$}

\item $\displaystyle\lim_{x\to\infty}\left[\sec\left(\displaystyle\frac{1}{x}\right)+1\right]$
\answer{$2$}

\item $\displaystyle\lim_{x\to-\infty}\displaystyle\sqrt{x^2+3x+5}+x$
\answer{$\displaystyle-\frac{3}{2}$}

\item $\displaystyle \lim_{x\to \infty }\frac{x^2+5x+4}{3x^2+2}$
\answer{$\displaystyle \frac{1}{3}$}

\item $\displaystyle \lim_{x\to \infty }\frac{4x^5-3x^2+6}{x^4+7}$
\answer{$\displaystyle \infty$}

\item $\displaystyle \lim_{x\to \infty }\frac{3x^4+6x-7}{x^5+10}$
\answer{$\displaystyle 0$}

\item $\displaystyle \lim_{x\to \infty }\frac{\sqrt{x^2+10}}{x+3}$
\answer{$\displaystyle 1$}

\item $\displaystyle \lim_{x\to -\infty }\frac{\sqrt{2x^2+3}}{x-2}$
\answer{$\displaystyle -\sqrt{2}$}

\item $\displaystyle \lim_{x\to -\infty }\frac{5x+\sqrt{x^2+1}}{x+5}$
\answer{$\displaystyle 4$}

\item $\displaystyle \lim_{x\to \infty }\left( \sqrt{x^2+4x+1}-x\right)$
\answer{$\displaystyle 2$}

\item $\displaystyle \lim_{x\to -\infty }\left(x+\sqrt{x^2-6x+5}\right)$
\answer{$\displaystyle 3$}

\end{problemblock}

\addtoanswerkey{\end{multicols}}

\lab{\clearpage}%
\begin{problemblock}[2]{Find the horizontal asymptotes of the following functions.}

\item $\displaystyle f(x)=\frac{3x+3}{2x-4}$
\answer{Horizontal asymptote: $y=\displaystyle\frac{3}{2}$}

\item $f(x)=x^3+5x+2$
\answer{No horizontal asymptotes}

\item $\displaystyle g(t)=\frac{\sqrt{t^2+3}}{t-2}$
\answer{Horizontal asymptotes: $y=-1$, $y=1$}

\item $\displaystyle f(x)=\frac{x^2-2x+1}{2x^2-2x-12}$
\answer{Horizontal asymptote: $\displaystyle y=\frac{1}{2}$}

\item $\displaystyle f(x)=\displaystyle\frac{\cos x}{x}$
\answer{Horizontal asymptote: $y=0$ (Hint: The Squeeze Theorem, generalized to a limit at infinity, can be used here to evaluate the limits.)}

\item $\displaystyle y=\frac{5x^2-3x+1}{x^2-16}$ \answer{Horizontal asymptote: $y=5$}

\item $f(x)=\displaystyle\frac{x^3+1}{x^3+x^2}$
\answer{Horizontal asymptote: $y=1$}

\item $\displaystyle F(x)=\frac{x}{\sqrt{4x^2+1}}$ \answer{Horizontal asymptotes: $\displaystyle y=-\frac{1}{2}$, $\displaystyle y=\frac{1}{2}$}

\end{problemblock}
