Limit of a function with powers (L'Hopital doesnt work) 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.

For finite $a>0,$
$$\lim_{x\to\infty}x^2(a^{1/x}-a^{1/(x+1)})=\lim_{x\to\infty}a^{1/(x+1)}\lim_{x\to\infty}x^2\left(a^{\{1/x-1/(x+1)\}}-1\}\right)$$
Now $\displaystyle x^2\left(a^{\{1/x-1/(x+1)\}}-1\}\right)=\dfrac{a^{\frac1{x(x+1)}}-1}{\frac1{x(x+1)}}\cdot\dfrac1{1+\frac1x}$
Set $\frac1{x(x+1)}=y$ to use $\lim_{y\to0}\dfrac{a^y-1}y=\ln a$
