TITLE: Limit of a function with powers (L'Hopital doesnt work)
QUESTION [5 upvotes]: I have a little problem with limit of this function:
$\lim_{x \to \infty} x^2(2017^{\frac{1}{x}} - 2017^{\frac{1}{x+1}})$
I have tried de L'Hopital rule twice, but it doesn't work. Now I have no idea how to do it.

REPLY [2 votes]: Since $\lim_{x\to\infty}2017^{1/(x+1)}=1$, your limit is
$$
\lim_{x\to\infty}2017^{1/(x+1)}x^2(2017^{1/(x^2+x)}-1)
=
\lim_{x\to\infty}x^2(2017^{1/(x^2+x)}-1)\\
$$
Now do $t^{-1}=x^2+x$, so
$$
x^2=\frac{t+2-\sqrt{t^2+4t}}{2t}
$$
and you get
$$
\lim_{t\to0^+}\frac{(2017^t-1)}{t}\frac{t+2-\sqrt{t^2+4t}}{2}=\ln2017
$$