TITLE: Help with a limit, function to the power of a function
QUESTION [3 upvotes]: 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 $y=\frac{1}{\sqrt{e}}$

REPLY [0 votes]: 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, 
$= \exp \left(\lim_{x\to\infty}\dfrac{x^3}{-2(x^3 +x^2)}\right)$
$= \dfrac 1{\sqrt e}$
{"set_name": "stack_exchange", "score": 3, "question_id": 2832958}