Evaluate $\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$ I am trying to evaluate
$$\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$$
for $a>0$.
The idea:
$$a^{1 \over x} = e^{\ln{a^{1 \over x}}}=e^{{1 \over x}\ln{a}} = 1 + {\ln{a} \over x} + {1 \over 2}{(\ln{a})^2 \over x^2}+{1 \over 6}{(\ln{a})^3 \over x^3}+\dots$$
$$a^{1 \over x}-a^{1 \over x+1} = e^{\ln{a^{1 \over x}}}-e^{\ln{a^{1 \over x+1}}}=e^{{1 \over x}\ln{a}}-e^{{1 \over x+1}\ln{a}} = \ln{a} \left({1\over x} - {1 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ {1\over x^2} - {1 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x^3} - {1 \over (x+1)^3}}\right)+\dots$$
$$x^2 \left(a^{1 \over x}-a^{1 \over x+1}\right) =\ln{a} \left(x - {x^2 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ 1 - {x^2 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x} - {x^2 \over (x+1)^3}}\right)+\dots$$
This is where I get stuck. I suppose the whole sum should converge to $\ln{a}$ but can't quite find the way to get there. I would appreciate any comments.
 A: You seem to have some issue with your Taylor expansion of $\exp$. Here, we will only need it to second or third order, but you forgot the linear term:
when $u\to 0$, we have $$e^u = 1+\boxed{\color{red} u} +\frac{u^2}{2} + o(u^2)$$
which is crucial here.
Using this, we get, since $\frac{\ln a}{x} \xrightarrow[x\to\infty]{}0$,
$$\begin{align}
a^{\frac{1}{x}} &= e^{\frac{\ln a}{x}}
= 1+\frac{\ln a}{x}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{1}\\a^{\frac{1}{x+1}}&= e^{\frac{\ln a}{x+1}}
= 1+\frac{\ln a}{x+1}+\frac{\ln^2 a}{(x+1)^2} + o\left(\frac{1}{x^2}\right) \\
&= 1+\frac{\ln a}{x}\left(1-\frac{1}{x} + o\left(\frac{1}{x}\right)  \right)+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \\
&= 1+\frac{\ln a}{x}-\frac{\ln a}{x^2}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{2}
\end{align}$$
using also that $\frac{1}{1+u} = 1-u+o(u)$ when $u\to 0$ (in the third line). Thus, the difference of the two will be
$$
a^{\frac{1}{x}} - a^{\frac{1}{x+1}} = \frac{\ln a}{x^2}  + o\left(\frac{1}{x^2}\right) 
$$
from (1) and (2) (the other terms cancel), and the limit will then be
$$
\lim_{x\to\infty}x^2\left( a^{\frac{1}{x}} - a^{\frac{1}{x+1}}\right) = \lim_{x\to\infty}x^2\left( \frac{\ln a}{x^2}  + o\left(\frac{1}{x^2}\right) \right) = \boxed{\ln a}
$$
as desired.
A: I got it with a different approach. Take the logarithm of the limit,
$\ln(x^2) + \ln(a^{1/x} - a^{1/(1+x)}) = \ln(x^2) + \frac{\ln(a)}{x} + \ln(1 - a^{1/(1+x)-1/x})$
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-e^{\frac{-\ln(a)}{x(1+x)}}) $
Since $\frac{-\ln(a)}{x(1+x)} \rightarrow 0$ as $x \rightarrow \infty$ we take a Taylor expansion of $e^{\frac{-\ln(a)}{x(1+x)}}$.
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-(1-\frac{\ln(a)}{x(1+x)} + \mathcal{O}(\frac{1}{x^4})) $
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(\frac{\ln(a)}{x(1+x)}) + \mathcal{O}(\frac{1}{x^2}) $ 
$\ln(\frac{x^2}{x(x+1)}) + \frac{\ln(a)}{x} + \ln(\ln(a)) + \mathcal{O}(\frac{1}{x^2}) \rightarrow \ln(\ln(a))$ as $x \rightarrow \infty$
Thus the limit itself is $\ln(a)$.
