Let $f$ be a continuous function on $\mathbb R$ such that $\lim_{x\to 0} f(\frac{1}{x})$ exists. Show that $f$ is uniformly continuous on $\mathbb R$.
My proof is as follows:
Let $n = \frac{1}{x}$ and $\lim_{r\to \infty} f(n)$ = $l$.
There exists a $M$ such that if $n \geq M$, $|f(n) - l|<\epsilon$.
Suppose $f$ is not uniformly continuous on $\mathbb R$.
Then for all $\epsilon > 0$, there exists a $\delta$ such that for some n, u belonging to R, $|n-u|<\delta$ $\implies$ $|f(n) - f(u)| \geq \epsilon$. Pick $u = M$, then f0r any $n$ such that $|n-u|<\delta$, we have $|f(n) - l| \geq \epsilon$. Contradiction.
Hence, $f$ is uniformly continuous.
I don't really know how to prove this. Can someone help please. Thanks