Using a Limit at Zero to Evaluate a Limit at Infinity

Question

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 :)

Answer 1

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.

Answer 2

For any function $f$ on $(0,\infty)$, $\lim_{x \to \infty} f(x)$ exists iff $\lim_{x \to 0+} f(\frac 1 x)$ exists and the limits are equal in this case.