<s>[INST] 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.
 [/INST] 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 $\ln(a)$.
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