Limits: if $ \lim_{x\to\infty}f(x) = 0 $, does $\lim_{x \to 0}f(1/x)$ exist?

Question

Let $f$ be a function: $$ f:\mathbb{R} \to \mathbb{R} $$ It is known that: $$ \lim_{x\to\infty}f(x) = 0 $$ I need to prove / disprove that the following limit exist: $$ \lim_{x \to 0}f\left(\frac{1}{x}\right) $$

Answer 1

Example:

$$f(x)=\begin{cases}\frac1x&;x>0\\1&;x\le0\end{cases}$$

$$\lim_{x\to\infty}f(x)=0$$

but

$$\lim_{x\to0^+}f(1/x)=\lim_{x\to0^+}x=0$$

$$\lim_{x\to0^-}f(1/x)=\lim_{x\to0^-}1=1$$

Answer 2

Maybe the simplest counterexample would be $f(x)=e^{-x}$.

The limit $$\lim_{x\to 0} e^{-\frac1x}$$ does not exist, because if $x$ is small but positive, $e^{-\frac1x}$ is close to $0$, but if $x$ is small and negative, then $e^{-\frac1x}$ is large.