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.
 A: 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 $\lim_{y\to0}\dfrac{a^y-1}y=\ln a$
A: 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.
A: 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
$$
\lim_{t\to0^+}\frac{(2017^t-1)}{t}\frac{t+2-\sqrt{t^2+4t}}{2}=\ln2017
$$
A: The best idea is probably to expand in power series (if you are allowed to do that). As Wolfram Alpha confirms, 
$$
2017^\frac{1}{x} - 2017^\frac{1}{x+1} = \frac{\log (2017)}{x^2} + o(1/x^2)
$$
as $x \to +\infty$, and therefore the limit equals $\log (2017)$.
