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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!

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Should this be the one side limit as x approaches 0? – Tyler Mar 29 '13 at 17:41
Let $f(x)=\frac{x}{|x|}$ Note that $\lim_{x\to\infty}f(x)=1$ while $\lim_{x\to 0} f(1/x)$ does not exist. – André Nicolas Mar 29 '13 at 17:58
@AndréNicolas: shall not it be then $\lim_{x\to {\color{red}+}\infty}f(x) = 1$? – Ilya Mar 29 '13 at 18:38
Ordinarily, at least in North American calculus texts, $\lim_{x\to\infty}$ and $\lim_{x\to +\infty}$ are synonyms, with the first more common. – André Nicolas Mar 29 '13 at 19:08
up vote 4 down vote accepted

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.

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