<s>[INST] 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.
 [/INST] make a change of variable $$x = \frac1{t^2}, \quad t = \frac1{\sqrt x }.$$ with that we have $$\begin{align}x\ln x + 2x\ln \sin (1/\sqrt x) &= \frac{1}{t^2} \ln \frac1{t^2} + \frac 2{t^2} \ln \sin t \\
&= \frac2{t^2}\left(\ln\sin \left(t\right) - \ln t  \right) =\frac2{t^2}\ln\left(\frac{\sin t}t\right) \\
&= \frac2{t^2}\ln\left(\frac{t - \frac16 t^3 + \cdots}t\right) =  \frac2{t^2}\ln\left(1 - \frac16 t^2 + \cdots\right) \\
&=\frac 2{t^2}\left(-\frac {t^2}6+\cdots\right)\\
&= -\frac13 \text{ as } t \to 0.\end{align}$$
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