Prove that for any $f(x)$, $$\lim_{x\to\infty}f(x)=\lim_{x\to 0}f\left(\frac{1}{x}\right).$$
I don't know for sure if this is true, I just thought about it. It seems very intuitive for a simple $f(x)$, but for more complex ones (such as the definition of $e$, where this does work) I wouldn't be sure if it always worked. Thanks!
The assumption is that for all $\epsilon>0$ there is a $N$ such that for $n\geq N$ you have $\lvert f(x) - L\rvert < \epsilon$ say.
Now Let $\epsilon>0$ be given. And choose $N$ so that for $x>N$ we indeed have $\lvert f(x) - L\rvert < \epsilon$. Now then that means that for $\frac{1}{x} < \frac{1}{N}$ we have $\lvert f(x) - L\rvert < \epsilon$. So for all $0<y< \frac{1}{N}$ (note that $y = \frac{1}{x}$ for some $x$) you have $\lvert f(y^{-1}) - L\rvert < \epsilon$.
So we have found a $\delta = \frac{1}{N}$ such that for all $0<y<\delta$ such that $\lvert f(y^{-1}) - L\rvert < \epsilon$. Hence
$$ \lim_{y\to 0^+} f(y^{-1}) = \lim_{x\to \infty} f(x) = L. $$ Here we have assumed that the limit is equal to a number. If the limit is infinity, then you can probably find an argument very similar to what I have written here.
