How to prove this limit's equality? Prove that $\lim\limits_{x\rightarrow 0^+}f(x)$=$\lim\limits_{x\rightarrow \infty}f(1/x)$.
I think that it could be solved with a variable change but I don't know how can I change te infitity to 0+.
 A: This looks obvious but it does need a sound proof. Before giving a formal $\epsilon, \delta$ type argument it is always better to have an informal argument and one also needs to understand that rigor is different from formalism.
So let us assume that $f(x) \to L$ as $x \to 0^{+}$. Now as $x \to 0^{+}$ we can see that $y = 1/x \to \infty$. And rewording it we see that as $y \to \infty$ we have $x = 1/y \to 0^{+}$ and hence $f(1/y) \to L$ as $y \to \infty$. Since the name of the variable is immaterial when calculating limits we can as well write $f(1/x) \to L$ as $x \to \infty$.
This can be easily translated into formal $\epsilon, \delta$ type argument. The statement $f(x) \to L$ as $x \to 0^{+}$ implies that for any $\epsilon > 0$ we have a number $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < x < \delta$.
We need to show that $f(1/x) \to L$ as $x \to \infty$. This would require us to show that corresponding to any $\epsilon > 0$ we must be able to find a number $N > 0$ such that $|f(1/x) - L| < \epsilon$ whenever $x > N$. To find $N$ we first choose a $\delta > 0$ corresponding to $\epsilon$ such that $|f(x) - L| < \epsilon$ whenever $0 < x < \delta$. Let $N = 1/\delta$ so that $N > 0$. Now consider a new variable $y$ such that $y > N$. This means that $0 < 1/y < \delta$ and hence $|f(1/y) - L| < \epsilon$. This shows that $f(1/y) \to L$ as $y \to \infty$ or changing the variable name $f(1/x) \to L$ as $x \to \infty$.
