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.

Let us consider $$A=x^2(a^{\frac{1}{x}} -a^{\frac{1}{x+1}})$$ $$a^{\frac{1}{x}}=e^{\frac{\log(a)}x}=1+\frac{\log (a)}{x}+\frac{\log ^2(a)}{2 x^2}+\frac{\log ^3(a)}{6
   x^3}+O\left(\frac{1}{x^4}\right)$$  Do the same for the other term; subtract from eash other, use common denominator and so on. 
For the first term,
$$a^{\frac{1}{x}} -a^{\frac{1}{x+1}}=\log(a)\left(\frac{1}{x}-\frac{1}{x+1} \right)+\cdots=\log(a)\frac{x+1-x}{x(x+1)}+\cdots=\frac{\log(a)}{x(x+1)}+\cdots$$ and so on.
You should arrive to $$a^{\frac{1}{x}} -a^{\frac{1}{x+1}}=\frac{\log (a)}{x^2}+\frac{\log ^2(a)-\log
   (a)}{x^3}+O\left(\frac{1}{x^4}\right)$$ whcih will show the limit and how it is approached.
