I think that this limit should be not defined

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

  • $\begingroup$ What did you try? $\endgroup$ – Ofek Gillon May 6 '17 at 10:23
  • 1
    $\begingroup$ Applying log properties then it must be not define $\endgroup$ – NIDHITANSH May 6 '17 at 10:26
  • $\begingroup$ 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. $\endgroup$ – John Hughes May 6 '17 at 10:35
  • $\begingroup$ When x multiply sinx1÷x^ and c tends to infinity so infinity×0 = not define $\endgroup$ – NIDHITANSH May 6 '17 at 10:38
  • $\begingroup$ Not. Go here. It may help :wolframalpha.com/input/… $\endgroup$ – The Dead Legend May 6 '17 at 10:44

Too many comments, no proposed answer so far: first ,substitute

$$y:=\frac1{\sqrt x}\;,\;\;\text{and observe that}\;\;x\to\infty\implies y\to 0\;,\;\;\text{so we get the limit}$$

$$\lim_{y\to0}\left(\frac1{y^2}\,\log\frac1{y^2}+\frac2{y^2}\,\log\sin y\right)=\lim_{y\to0}\frac{-2\log y+2\log\sin y}{y^2}\stackrel{l'H}=\lim_{y\to0}\frac{-\frac2y+\frac2{\sin y}\cdot\cos y}{2y}=$$$${}$$

$$=\lim_{y\to0}\frac{-\sin y+y\cos y}{y^2\sin y}\stackrel{l'H}=\lim_{y\to0}\frac{-y\sin y}{2y\sin y+y^2\cos y}=\lim_{y\to0}\frac{-\sin y}{2\sin y+y\cos y}\stackrel{l'H}=$$$${}$$

$$=\lim_{y\to0}\frac{-\cos y}{3\cos y-y\sin y}=-\frac13$$

  • $\begingroup$ Thank u sir. But it also be sloved by expansions $\endgroup$ – NIDHITANSH May 6 '17 at 11:30
  • $\begingroup$ @NIDHITANSH And perhaps also in other ways. The above is just one of them, and (Taylor or power series) expansions seem TO ME way messier than this. But it may be a matter of taste. $\endgroup$ – DonAntonio May 6 '17 at 11:33
  • $\begingroup$ It's very good method too. $\endgroup$ – NIDHITANSH May 6 '17 at 11:34

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