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\noindent {\bf Question}

\noindent Evaluate the following limits.
\begin{enumerate}
\item $\lim_{x\to 2} (1-\cos(\pi x)) / \sin^2(\pi x)$;
\item $\lim_{x\to {-1}} (x^7+1) / (x^3+1)$;
\item $\lim_{x\to 3} (1+\cos(\pi x)) / \tan^2(\pi x)$;
\item $\lim_{x\rightarrow 1} (1-x+\ln(x))/(1+\cos(\pi x))$;
\item $\lim_{x\rightarrow\infty} (\ln(x))^{1/x}$;
\item $\lim_{x\rightarrow 2} (x^2+x-6)/(x^2-4)$;
\item $\lim_{x\rightarrow 0} (x+\sin(2x))/(x-\sin(2x))$;
\item $\lim_{x\rightarrow 0} (e^x-1)/x^2$;
\item $\lim_{x\rightarrow 0} (e^x+e^{-x}-x^2-2)/(\sin^2(x)-x^2)$;
\item $\lim_{x\rightarrow\infty} \ln(x)/x$;
\item $\lim_{x\rightarrow 2} (x^3-x^2-x-2)/(x^3-3x^2+3x-2)$;
\item $\lim_{x\rightarrow 1} (x^3-x^2-x+1)/(x^3-2x^2+x)$;
\end{enumerate}


\medskip

\noindent {\bf Answer}

\noindent [Note that for some of these limits, we do not need to
use as heavy a piece of machinery as l'Hopital's rule, just some
clever simplifying.]
\begin{enumerate}
\item since this limit has the indeterminate form $\frac{0}{0}$ (since
both $\lim_{x\rightarrow 2} (1-\cos(\pi x)) =0$ and
$\lim_{x\rightarrow 2} \sin^2(\pi x) =0$), we may use l'Hopital's
rule:
\[ \lim_{x\rightarrow 2} \frac{1-\cos(\pi x)}{\sin^2(\pi x)}
=\lim_{x\rightarrow 2} \frac{\pi \sin(\pi x)}{2\pi\sin(\pi
x)\cos(\pi x)} = \lim_{x\rightarrow 2} \frac{1}{2\cos(\pi x)}
=\frac{1}{2}. \]

\medskip
\noindent (Note that we may also evaluate this limit without
l'Hopital's rule, using the trigonometric identity $\sin^2(\theta)
+\cos^2(\theta) =1$, as follows:
\[ \lim_{x\rightarrow 2} \frac{1-\cos(\pi x)}{\sin^2(\pi x)} =
\lim_{x\rightarrow 2} \frac{1-\cos(\pi x)}{1 -\cos^2(\pi x)} =
\lim_{x\rightarrow 2} \frac{1}{1 +\cos(\pi x)} = \frac{1}{2}.
\left. \right) \]
\item again, here we have the choice of factoring or using l'Hopital's
rule.  I feel like factoring:
\begin{eqnarray*}
\lim_{x\rightarrow {-1}} \frac{x^7+1}{x^3+1} & = &
\lim_{x\rightarrow {-1}} \frac{(x+1)(x^6 -x^5 +x^4 -x^3 +x^2 -x
+1)}{(x+1)(x^2 -x+1)} \\
 & = & \lim_{x\rightarrow {-1}} \frac{x^6 -x^5 +x^4 -x^3 +x^2 -x
+1}{x^2 -x+1} = \frac{7}{3}. \end{eqnarray*}
\item write $\tan(z) = \sin(z) /\cos(z)$ and simplify:
\begin{eqnarray*} \lim_{x\rightarrow 3} \frac{1+\cos(\pi
x)}{\tan^2(\pi x)} & = & \lim_{x\rightarrow 3} \frac{(1+\cos(\pi
x))\cos^2(\pi x)}{\sin^2(\pi x)} \\
 & = & \lim_{x\rightarrow 3} \frac{(1+\cos(\pi x))\cos^2(\pi x)}{1
-\cos^2(\pi x)} = \lim_{x\rightarrow 3} \frac{\cos^2(\pi x)}{1
-\cos(\pi x)} = \frac{1}{2}. \end{eqnarray*}
\item as this has the indeterminate form $\frac{0}{0}$, and since
there seems to be no easy simplification possible, we use
l'Hopital's rule:
\[ \lim_{x\rightarrow 1} \frac{1-x+\ln(x)}{1+\cos(\pi x)}
=\lim_{x\rightarrow 1} \frac{ -1 +\frac{1}{x}}{-\pi \sin(\pi x)}.
\] Since this limit still has the indeterminate form
$\frac{0}{0}$, we may use l'Hopital's rule again:
\[ \lim_{x\rightarrow 1} \frac{ -1 +\frac{1}{x}}{-\pi \sin(\pi x)} =
\lim_{x\rightarrow 1} \frac{ -\frac{1}{x^2}}{-\pi^2 \cos(\pi x)} =
-\frac{1}{\pi^2}. \]
\item this has the indeterminate form $\infty^0$, and so we rewrite
it:
\[ \lim_{x\rightarrow\infty} (\ln(x))^{1/x} =\lim_{x\rightarrow\infty}
\left( e^{\ln(\ln(x))}\right)^{1/x} = e^{\lim_{x\rightarrow\infty}
\ln(\ln(x))/x}. \] The exponent has the indeterminate form
$\frac{\infty}{\infty}$, and so we may use l'Hopital's rule:
\[ \lim_{x\rightarrow\infty} \frac{\ln(\ln(x))}{x} =
\lim_{x\rightarrow\infty} \frac{\frac{1}{\ln(x)}\cdot
\frac{1}{x}}{1} = 0. \] Hence, we see that
\[ \lim_{x\rightarrow\infty} (\ln(x))^{1/x} =
e^{\lim_{x\rightarrow\infty} \ln(\ln(x))/x} =e^0 =1. \]
\item factoring, we see that
\[ \lim_{x\rightarrow 2} \frac{x^2+x-6}{x^2-4} =\lim_{x\rightarrow 2}
\frac{(x-2)(x+3)}{(x-2)(x+2)} = \lim_{x\rightarrow 2}
\frac{x+3}{x+2} =\frac{5}{4}. \]
\item as this limit has the indeterminate form $\frac{0}{0}$, we may
use l'Hopital's rule:
\[ \lim_{x\rightarrow 0} \frac{x+\sin(2x)}{x-\sin(2x)} =
\lim_{x\rightarrow 0} \frac{1+ 2\cos(2x)}{1 -2\cos(2x)} = \frac{1
+2}{1-2} = -3. \]
\item since this limit has the indeterminate form
$\frac{\infty}{\infty}$, we may apply l'Hopital's rule:
\[ \lim_{x\rightarrow 0} \frac{e^x-1}{x^2} = \lim_{x\rightarrow 0}
\frac{e^x}{2 x} = \lim_{x\rightarrow 0} \frac{e^x}{2} = \infty. \]
(The second equality follows from applying l'Hopital's rule a
second time, which is valid since the limit still has the
indeterminate form $\frac{\infty}{\infty}$.)
\item in this limit, though we need to check at each stage, we will
apply l'Hopital's rule four times, as the original limit has the
indeterminate form $\frac{0}{0}$, and each of the first three
applications of l'Hopital's rule results in a limit still in the
indeterminate form $\frac{0}{0}$.
\begin{eqnarray*}
\lim_{x\rightarrow 0} \frac{e^x+e^{-x}-x^2-2}{\sin^2(x)-x^2} =
\lim_{x\rightarrow 0} \frac{e^x -e^{-x}-2x}{2\sin(x)\cos(x)-2x} &
= & \lim_{x\rightarrow 0} \frac{e^x -e^{-x}-2x}{\sin(2x) -2x} \\
 & = & \lim_{x\rightarrow 0} \frac{e^x +e^{-x}-2}{2\cos(2x) -2} \\
 & = & \lim_{x\rightarrow 0} \frac{e^x -e^{-x}}{-4\sin(2x)} \\
 & = & \lim_{x\rightarrow 0} \frac{e^x +e^{-x}}{-8\cos(2x)} =
-\frac{1}{4}. \end{eqnarray*}
\item this limit has the indeterminate form $\frac{\infty}{\infty}$,
and so we apply l'Hopital's rule:
\[ \lim_{x\rightarrow\infty} \frac{\ln(x)}{x} =
\lim_{x\rightarrow\infty} \frac{\frac{1}{x}}{1} = 0. \]
\item here, we first attempt to evaluate the limit by factoring, a
sensible first step for limits of rational functions:
\[ \lim_{x\rightarrow 2} \frac{x^3-x^2-x-2}{x^3-3x^2+3x-2}
=\lim_{x\rightarrow 2} \frac{(x-2)(x^2 +x+1)}{(x-2)(x^2 -x+1)}
=\lim_{x\rightarrow 2} \frac{x^2 +x+1}{x^2 -x+1} = \frac{7}{3}. \]
\item again, we first attempt to evaluate the limit by factoring:
\[ \lim_{x\rightarrow 1} \frac{x^3-x^2-x+1}{x^3-2x^2+x}
=\lim_{x\rightarrow 1} \frac{(x-1)(x^2 -1)}{x(x-1)^2} =
\lim_{x\rightarrow 1} \frac{x+1}{x} = 2.\]
\end{enumerate}


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