# Limits, continuities & differentiability

I think that this limit should be not defined

$$\lim_{x\rightarrow \infty}x \ln x+2x\ln \sin \left(\frac{1}{\sqrt{x}} \right)$$

• What did you try? – Ofek Gillon May 6 '17 at 10:23
• Applying log properties then it must be not define – NIDHITANSH May 6 '17 at 10:26
• Now: tell us what you've tried, and why you think that the limit does not exist, and that'll help us know how much you understand, so we can better help you. – John Hughes May 6 '17 at 10:35
• When x multiply sinx1÷x^ and c tends to infinity so infinity×0 = not define – NIDHITANSH May 6 '17 at 10:38
• Not. Go here. It may help :wolframalpha.com/input/… – The Dead Legend May 6 '17 at 10:44

$$y:=\frac1{\sqrt x}\;,\;\;\text{and observe that}\;\;x\to\infty\implies y\to 0\;,\;\;\text{so we get the limit}$$
$$\lim_{y\to0}\left(\frac1{y^2}\,\log\frac1{y^2}+\frac2{y^2}\,\log\sin y\right)=\lim_{y\to0}\frac{-2\log y+2\log\sin y}{y^2}\stackrel{l'H}=\lim_{y\to0}\frac{-\frac2y+\frac2{\sin y}\cdot\cos y}{2y}=$$$${}$$
$$=\lim_{y\to0}\frac{-\sin y+y\cos y}{y^2\sin y}\stackrel{l'H}=\lim_{y\to0}\frac{-y\sin y}{2y\sin y+y^2\cos y}=\lim_{y\to0}\frac{-\sin y}{2\sin y+y\cos y}\stackrel{l'H}=$$$${}$$
$$=\lim_{y\to0}\frac{-\cos y}{3\cos y-y\sin y}=-\frac13$$