Find the limit $ L = \lim\limits_{x \to \infty} \left( x \ln x + 2 x \ln \sin \frac{1}{\sqrt{x}} \right) $. 
Problem: Find the limit $ \displaystyle L = \lim_{x \to \infty} \left( x \ln x + 2 x \ln \sin \frac{1}{\sqrt{x}} \right) $.

Please suggest how to proceed in this problem. I will be grateful to you. Thanks.
 A: make a change of variable $$x = \frac1{t^2}, \quad t = \frac1{\sqrt x }.$$ with that we have $$\begin{align}x\ln x + 2x\ln \sin (1/\sqrt x) &= \frac{1}{t^2} \ln \frac1{t^2} + \frac 2{t^2} \ln \sin t \\
&= \frac2{t^2}\left(\ln\sin \left(t\right) - \ln t  \right) =\frac2{t^2}\ln\left(\frac{\sin t}t\right) \\
&= \frac2{t^2}\ln\left(\frac{t - \frac16 t^3 + \cdots}t\right) =  \frac2{t^2}\ln\left(1 - \frac16 t^2 + \cdots\right) \\
&=\frac 2{t^2}\left(-\frac {t^2}6+\cdots\right)\\
&= -\frac13 \text{ as } t \to 0.\end{align}$$
A: First,
look at the two terms separately.
Then,
for the second term,
use the result that,
for small $x$,
$\sin(x) \approx x- x^3/6$.
And, of course,
$\ln \sqrt{x}
=\frac12 \ln x
$.
After using these,
recombine and
see what cancels.
A: Hint:
$$ x\ln x +2x\ln \sin\frac{1}{\sqrt{x}}=x \ln \left( x  \sin^2 \frac{1}{\sqrt{x}} \right)=\frac{ln \left( x  \sin^2 \frac{1}{\sqrt{x}} \right)}{\frac{1}{x}}$$
Use L'H.
A: We have
$$\sin(1/\sqrt{x}) = \dfrac1{\sqrt{x}} - \dfrac1{3!} \dfrac1{x\sqrt{x}} + \mathcal{O}\left(1/x^{5/2} \right)$$
We have
\begin{align}
\ln(\sin(1/\sqrt{x})) & = \ln\left(\dfrac1{\sqrt{x}} - \dfrac1{3!} \dfrac1{x\sqrt{x}} + \mathcal{O}\left(1/x^{5/2} \right)\right)\\
& = \ln\left(1/\sqrt{x} \right) + \ln\left(1-1/(6x) + \mathcal{O}(1/x^2)\right) = -\dfrac{\ln(x)}2 - \dfrac1{6x} + \mathcal{O}(1/x^2)
\end{align}
Hence, we have
$$x\ln(x)+2x\ln(\sin(1/\sqrt{x})) = x\ln(x) - x\ln(x) - \dfrac13 + \mathcal{O}(1/x) = - \dfrac13 + \mathcal{O}(1/x)$$
Hence, as $x \to \infty$, the limit is $-\dfrac13$.
