Prove $\lim_{x \to \infty} f(x) = L \iff \lim_{x \to 0} g(x) = L$ Let $f(x) = g(1/x)$ for $x>0$. Prove: $\lim_{x \to \infty} f(x) = L \iff \lim_{x \to 0} g(x) = L$ for some $L \in \mathbb{R}$. 
I assume I am supposed to use l'Hopital's rule in some way (considering that is what section we are in). I've tried looking at the definition of a limit and the sequential criterion for limits, but I have no idea where to go.
Just a push in the right direction would be awesome. Thanks in advance.

Here is the right direction:
Given $\lim_{x \rightarrow \infty} f(x) = L$ we know $\exists N : x> N \Rightarrow |f(x) -L| < \epsilon$.
Set $x' = \frac 1 x$ and $\delta = \frac 1 N$. Then $x>N \Leftarrow\Rightarrow x' < \frac 1 N = \delta$.
Thus $x' < \delta \Rightarrow x> N \Rightarrow |f(x) -L| = |f(\frac 1 {x'}) -L| = |g(x') -L| < \epsilon$. And so $\lim_{x' \rightarrow 0}g(x') = L$.
Note: We are actually assuming $x'>0$, so the limit only works as $x'$ approaches $0$ from the right.
