Help with a limit, function to the power of a function

Question

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 $y=\frac{1}{\sqrt{e}}$

Answer 1

$$=\frac{\sin\frac{1}{x}}{\frac{1}{x}}\frac{\ln(x\ln(1+\frac{1}{x}))}{x\ln(1+\frac{1}{x})-1}x(x\ln(1+\frac{1}{x})-1)$$ the first two fractions are standard limits, now apply Hopital twice to

$$\frac{x\ln(1+\frac{1}{x})-1}{\frac{1}{x}}$$

Answer 2

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 $$\frac1{\sqrt{e}}\left(1 + \frac{5 z}{24} - \frac{23 z^2}{1152} + \frac{18769 z^3}{414720} - \frac{1526443 z^4}{39813120} + O\left(z^5\right)\right). $$

Answer 3

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

$$\left(x\ln\left(1+\frac{1}{x}\right)\right)^{x^2\sin(1/x)}=\left(1-\frac1{2x}+o\left(\frac1{x}\right)\right)^{x^2\sin(1/x)}=\left[\left(1-\frac1{2x}+o\left(\frac1{x}\right)\right)^{\frac1{\frac1{-2x}+o\left(\frac1{x}\right)}}\right]^{\frac{\sin \frac1x}{\frac1x}\left(-\frac1{2}+o\left(1\right)\right)}\to e^{-\frac12}$$

Answer 4

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,

$= \exp \left(\lim_{x\to\infty}\dfrac{x^3}{-2(x^3 +x^2)}\right)$

$= \dfrac 1{\sqrt e}$