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\noindent {\bf Question}

\noindent For each of the functions given below, determine whether
or not $\lim_{x\to 0} f(x)$ exists; if the limit does exist,
determine its value wherever possible.
\begin{enumerate}
\item $f(x)=\sin(x)\sin(\frac{1}{x})$, for $x\ne 0$;
\item $f(x)=\cos(x)$ for $x\ne 0$, and $f(0)=2$;
\item $f(x)=[3x+1]$ (where $[x]$ is the largest integer or floor
function);
\item $f(x)=\sin(\sin(\frac{1}{x}))$, for $x\ne 0$;
\item $f(x)=\cos(x)$, if $x$ is a positive rational multiple of
$\pi$, and $f(x) =1$ otherwise;
\item $f(x)= \frac{\sin(x)}{|x|}$ for $x\ne 0$;
\end{enumerate}

\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item use the squeeze law.  We have that $-1\le \sin(\frac{1}{x} )\le
1$ for all $x\ne 0$, and that $\lim_{x\rightarrow 0} \sin(x) =0$.
So, we can bound $f(x)$ below by $-\sin(x)$ and above by
$\sin(x)$.  Since $\lim_{x\rightarrow 0} -\sin(x) =
\lim_{x\rightarrow 0} \sin(x) = 0$, we have that
$\lim_{x\rightarrow 0} \sin(x)\sin( \frac{1}{x} ) =0$. [Note that
the fact that $f(x)$ is not defined at $0$ does not matter, since
evaluating $\lim_{x\rightarrow 0} f(x)$ depends only on what's
happening with $f(x)$ near $0$, and not at all on what's happening
at $0$.]
\item since $\lim_{x\rightarrow 0} \cos(x) = 1$, and since
$f(x)=\cos(x)$ except at $0$, we have that $\lim_{x\rightarrow 0}
f(x) = \lim_{x\rightarrow 0} \cos(x) =1$.  [This is another
reflection of the fact that $\lim_{x\rightarrow 0} f(x)$ does not
care about the value of $f(x)$ at $0$, but only on the values of
$f(x)$ near $0$.]
\item note that $f(x) = 0$ for $-\frac{1}{3} <x \le 0$, and so
$\lim_{x\rightarrow 0-} f(x) = 0$.  Also, $f(x) = 1$ for $0 <x <
\frac{1}{3}$, and so $\lim_{x\rightarrow 0+} f(x) = 1$.  Since
$\lim_{x\rightarrow 0+} f(x) \ne \lim_{x\rightarrow 0-} f(x)$, we
see that $\lim_{x\rightarrow 0} f(x)$ does not exist.
\item as $x\rightarrow 0+$, we see that $\frac{1}{x}\rightarrow
\infty$, and so $\sin( \frac{1}{x} )$ oscillates between $-1$ and
$1$. Hence, as $x\rightarrow 0+$, we have that $f(x)$ oscillates
between $\sin(-1)$ and $\sin(1)$, and so $\lim_{x\rightarrow 0+}
f(x)$ does not exist.  Hence, $\lim_{x\rightarrow 0} f(x)$ does
not exist.
\item we apply the squeeze rule, since $\cos(x)\le f(x)\le 1$ for all
$x$ near $0$.  Since both $\lim_{x\rightarrow 0} \cos(x)= 1$ and
$\lim_{x\rightarrow 0} 1 =1$, we have that $\lim_{x\rightarrow 0}
f(x) =1$.
\item to evaluate this limit, we recall from calculus that
$\lim_{x\rightarrow 0} \frac{\sin(x)}{x} =1$, and so by the lemma
below, we have that $\lim_{x\rightarrow 0+} \frac{\sin(x)}{x} =
\lim_{x\rightarrow 0-} \frac{\sin(x)}{x} =1$.

Lemma:

$\lim_{x\rightarrow a} f(x) =L$ if and only if $\lim_{x\rightarrow
a+} f(x) = \lim_{x\rightarrow a-} f(x) = L$.

\medskip
\noindent For $x >0$, we have that $|x| =x$, and so
$\lim_{x\rightarrow 0+} \frac{\sin(x)}{|x|} =\lim_{x\rightarrow
0+} \frac{\sin(x)}{x} = 1$. However, for $x <0$, we have that $|x|
=-x$, and so $\lim_{x\rightarrow 0-} \frac{\sin(x)}{|x|} =
-\lim_{x\rightarrow 0-} \frac{\sin(x)}{x} = -1$.  Since
$\lim_{x\rightarrow 0+} \frac{\sin(x)}{|x|} \ne \lim_{x\rightarrow
0-} \frac{\sin(x)}{|x|}$, we see that $\lim_{x\rightarrow 0}
\frac{\sin(x)}{|x|}$ does not exist.
\end{enumerate}


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