Q:

Using a Limit at Zero to Evaluate a Limit at Infinity

When does the following hold?
$\lim_{x \rightarrow \infty} f(x)=\lim_{x \rightarrow 0^+} f(\frac{1}{x})$
Why? Assume the right-hand side limit exists. An example of such a function would be $f(x)=\frac{1}{x}$.
Motivation: I would like to show that 
$\lim_{x \rightarrow \infty} x \sin (\frac{1}{x})=1$. 
If I can justify using 
$\lim_{x \rightarrow \infty} x\sin (\frac{1}{x})=\lim_{x \rightarrow 0^+} \frac{\sin (x)}{x}$,
this will yield the desired result as 
$\lim_{x \rightarrow 0^+} \frac{\sin (x)}{x}=1$.
Thank you :)

A:

Here is my proof that 
$\lim_{x \rightarrow \infty} f(x) =L$
if and only if
$\lim_{x \rightarrow 0^+} f(\frac{1}{x})=L$.
We begin by showing that if $\lim_{x \rightarrow \infty} f(x) =L$ then $\lim_{x \rightarrow 0^+} f(\frac{1}{x})=L$.
Correspondingly, suppose $\lim_{x \rightarrow \infty} f(x) =L$.
Let $\epsilon > 0$. Choose $N > 0$ such that if $x>N>0$, then $|f(x)-L|<\epsilon$ (here we are making use of the definition of a limit at infinity). Note that $x>N>0$ is equivalent to $0<\frac{1}{x}<\frac{1}{N}$. Let $u=\frac{1}{x}$ and $\delta = \frac{1}{N}$. It follows that if $0<u<\delta$, then $|f(\frac{1}{u})-L|<\epsilon$. As $\epsilon>0$ was arbitrary, it follows that $\lim_{x \rightarrow 0^+} f(\frac{1}{x})=L$.
We will now show that if $\lim_{x \rightarrow 0^+} f(\frac{1}{x})=L$ then $\lim_{x \rightarrow \infty} f(x) =L$.
Correspondingly, suppose that $\lim_{x \rightarrow 0^+} f(\frac{1}{x})=L$. 
Let $\epsilon >0$. Choose $\delta>0$ such that if $0<x<\delta$, then $|f(\frac{1}{x})-L|<\epsilon$ (here we are making use of the definition of a limit). Note that $0<x<\delta$ is equivalent to $\frac{1}{x}>\frac{1}{\delta}>0$. Let $u=\frac{1}{x}$ and $N=\frac{1}{\delta}$. It follows that if $u>N>0$, then $|f(u)-L|<\epsilon$. As $\epsilon >0$ was arbitrary, it follows that $\lim_{x \rightarrow \infty} f(x)=L$.
This completes the proof.

