Limits and sequences question Let 
$$a_n=n^x(n^{1/n^2}−1).$$ 
Show that 
$$\lim_{n \to \infty} \frac{a_n}{\ln(n)/(n^{2-x})} = 1. $$ 
It is on the study guide for my final exam, which is tomorrow so I am trying to figure it out. Thanks
I really can't figure this out
 A: Let $x=1/n^2$ so that $x \to 0^+$.
Then, we have $n=x^{-1/2}$ whereupon substitution yields
$$\begin{align}
\lim_{n\to \infty}\frac{n^2(n^{1/n^2}-1)}{\log n}&=\lim_{x\to 0^+}\frac{x^{-1}(x^{-x/2}-1)}{\log(x^{-1/2})}\\\\
&=-2\lim_{x\to 0^+}\frac{(x^{-x/2}-1)}{x\log x}
\end{align}$$
Note that both the numerator and denominator go to zero as $x\to 0$.  Thus, we can apply L'Hospital's Rule.  To that end, we have
$$\begin{align}
-2\lim_{x\to 0^+}\frac{(x^{-x/2}-1)}{x\log x}&=-2\lim_{x\to 0^+}\frac{-\frac12 x^{-x/2}(1+\log x)}{1+\log x}\\\\
&=\lim_{x\to 0^+} x^{-x/2}\\\\
&=1
\end{align}$$
as was to be shown!!
A: Solution for an arbitrary $x$:
First we plug $a_n=n^x(n^{1/n^2}−1)$ into $\lim\limits_{n \to \infty} \frac{a_n}{\ln(n)/(n^{2-x})} $ and get rid of $x$ due to $\ln(n)/(n^{2-x})=\ln(n)\cdot n^{x-2}$:
$$\lim\limits_{n \to \infty} \frac{a_n}{\ln(n)/(n^{2-x})} =
\lim\limits_{n \to \infty} \frac{n^x(n^{1/n^2}−1)}{\ln(n)/(n^{2-x})} =
\lim\limits_{n \to \infty} \frac{n^x(n^{1/n^2}−1)}{\ln(n)\cdot n^{x-2}} =
\lim\limits_{n \to \infty} \frac{n^{1/n^2}−1}{\ln(n)\cdot \frac{1}{n^2}} ;$$
$n^{1/n^2}$ is exactly $e^{\ln(n)\cdot \frac{1}{n^2}}$, so let $b_n=\ln(n)\cdot \frac{1}{n^2}$ and $f(x)=\frac{e^x-1}{x}$, we have $\lim\limits_{n\to\infty} b_n=0$ and $\lim\limits_{n\to\infty} f(b_n)=\lim\limits_{x\to 0}f(x)$; $\lim\limits_{x\to 0}\frac{e^x-1}{x}=1$.
