Limit $\lim_{x\rightarrow \infty}x \ln x+2x\ln \sin \left(\frac{1}{\sqrt{x}} \right)$ I think that this limit should be not defined 
$$\lim_{x\rightarrow  \infty}x \ln x+2x\ln \sin \left(\frac{1}{\sqrt{x}} \right)$$
 A: Too many comments, no proposed answer so far: first ,substitute
$$y:=\frac1{\sqrt x}\;,\;\;\text{and observe that}\;\;x\to\infty\implies y\to 0\;,\;\;\text{so we get the limit}$$
$$\lim_{y\to0}\left(\frac1{y^2}\,\log\frac1{y^2}+\frac2{y^2}\,\log\sin y\right)=\lim_{y\to0}\frac{-2\log y+2\log\sin y}{y^2}\stackrel{l'H}=\lim_{y\to0}\frac{-\frac2y+\frac2{\sin y}\cdot\cos y}{2y}=$$$${}$$
$$=\lim_{y\to0}\frac{-\sin y+y\cos y}{y^2\sin y}\stackrel{l'H}=\lim_{y\to0}\frac{-y\sin y}{2y\sin y+y^2\cos y}=\lim_{y\to0}\frac{-\sin y}{2\sin y+y\cos y}\stackrel{l'H}=$$$${}$$
$$=\lim_{y\to0}\frac{-\cos y}{3\cos y-y\sin y}=-\frac13$$
A: You can do it without substitution,
$$\lim_{x\to \infty}\left(x\ln x+2x\ln\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$
$$=\lim_{x\to \infty}2x\left(\ln\sqrt{x}+\ln\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$
$$=\lim_{x\to \infty}2x\ln\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$
$$=\lim_{x\to \infty}2x\ln\left(\sqrt{x}\left(\frac{(1/\sqrt{x})}{1!}-\frac{(1/\sqrt{x})^3}{3!}+\frac{(1/\sqrt{x})^5}{5!}+\ldots\right)\right)$$
$$=\lim_{x\to \infty}2x\ln\left(1-\frac{1}{6x}+\frac{1}{120x^2}-\ldots\right)$$
$$=\frac13\lim_{x\to \infty}\frac{\ln\left(1-\frac{1}{6x}\left(1+\frac{1}{20x}-\ldots\right)\right)}{\frac{1}{6x}\left(1+\frac{1}{20x}-\ldots\right)}\cdot \left(1+\frac{1}{20x}-\ldots\right)$$
$$=\frac{1}{3}(-1)\cdot 1$$$$=\color{blue}{-\frac13}$$
