Help with a limit, function to the power of a function I have the following limit:
$$y=\lim_{x\to\infty}
      \left(x\ln\left(1+\frac{1}{x}\right)\right)^{x^2\sin(1/x)}$$
From here I do the following:
$$\ln(y)=\lim_{x\to\infty}x^2\sin(\frac{1}{x})ln(x\ln(1+\frac{1}{x}))=\lim_{x\to\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x^2}}\ln(\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}})$$
And from here on I'm stuck with no obvious way to apply L'hopital's rule. Any tips? According to limit calculators $\ln(y)$ should have a value of $-\frac{1}{2}$ and then $y=\frac{1}{\sqrt{e}}$
 A: $$=\frac{\sin\frac{1}{x}}{\frac{1}{x}}\frac{\ln(x\ln(1+\frac{1}{x}))}{x\ln(1+\frac{1}{x})-1}x(x\ln(1+\frac{1}{x})-1)$$ the first two fractions are standard limits, now apply Hopital twice to 
$$\frac{x\ln(1+\frac{1}{x})-1}{\frac{1}{x}}$$
A: Letting $z = 1/x$,
$$
\begin{split}
L &= \lim_{x\to\infty}(x\ln(1+1/x))^{x^2\sin(1/x)}\\
&=\lim_{z\to 0}(\ln(1+z)/z)^{\sin(z)/z^2}\\
&=\lim_{z\to 0}((z-z^2/2+O(z^3))/z)^{(z-z^3/6+O(z^5))/z^2}\\
&=\lim_{z\to 0}(1-z/2+O(z^2))^{(1-z^2/6+O(z^4))/z}\\
&=\lim_{z\to 0}\exp(\ln(1-z/2+O(z^2))(1-z^2/6+O(z^4))/z)\\
&=\lim_{z\to 0}\exp((-z/2+O(z^2))(1-z^2/6+O(z^4))/z)\\
&=\lim_{z\to 0}\exp((-1/2+O(z))(1-z^2/6+O(z^4)))\\
&=\lim_{z\to 0}\exp(-1/2+O(z))\\
&=e^{-1/2}\\
\end{split}
$$
Note:
Wolfy agrees
and says that the expansion is
$$\frac1{\sqrt{e}}\left(1 + \frac{5 z}{24} - \frac{23 z^2}{1152}
                     + \frac{18769 z^3}{414720} - \frac{1526443 z^4}{39813120}
                     + O\left(z^5\right)\right).
$$
A: We have that by Taylor's expansion
$$\ln\left(1+\frac{1}{x}\right)=\frac1x-\frac1{2x^2}+o\left(\frac1{x^2}\right)\implies x\ln\left(1+\frac{1}{x}\right)=1-\frac1{2x}+o\left(\frac1{x}\right)$$
then
$$\left(x\ln\left(1+\frac{1}{x}\right)\right)^{x^2\sin(1/x)}=\left(1-\frac1{2x}+o\left(\frac1{x}\right)\right)^{x^2\sin(1/x)}=\left[\left(1-\frac1{2x}+o\left(\frac1{x}\right)\right)^{\frac1{\frac1{-2x}+o\left(\frac1{x}\right)}}\right]^{\frac{\sin \frac1x}{\frac1x}\left(-\frac1{2}+o\left(1\right)\right)}\to e^{-\frac12}$$
A: Theorem: Let $c_n \to L$ where $L \ne 0$. Then, then, $\displaystyle \lim_{n \to \infty} b_n c_n$ exists iff $\displaystyle \lim_{n \to \infty} b_n$ exists, and the values (if they exist) correspond in the usual way, i.e. $\displaystyle \lim_{n \to \infty} b_n c_n = L \lim_{n \to \infty} b_n$
$y=\lim_{x\to\infty}
      \left(x\ln\left(1+\frac{1}{x}\right)\right)^{x^2\sin(1/x)}$
$= e^{\lim_{x \to \infty}x.\color{blue} {x\sin \dfrac 1x} \ln (x \ln (1+\dfrac1x))}$
$= \exp(\lim_{x\to\infty}x \dfrac{(\ln(x\ln\left(1+\dfrac 1x \right)\color{red}{-1+1})}{\color{blue}{x \ln {(1+\dfrac 1x)-1}}}\color{blue}{(x\ln \left(1+\dfrac 1 x\right)-1)}$
$= \exp (\lim_{x\to\infty}x^2\left(\ln\left(1+\dfrac 1x\right)-\dfrac 1 x\right)$
$= \exp\left(\lim_{x\to\infty}\dfrac{\ln{(1+\frac1x)-\frac1x}}{\dfrac 1 {x^2}}\right)$
Now apply L Hopital's rule once, 
$= \exp \left(\lim_{x\to\infty}\dfrac{x^3}{-2(x^3 +x^2)}\right)$
$= \dfrac 1{\sqrt e}$
