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

Question

I think that this limit should be not defined

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

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 06 '17 at 10:35
When x multiply sinx1÷x^ and c tends to infinity so infinity×0 = not define — NIDHITANSH, May 06 '17 at 10:38
Not. Go here. It may help :http://www.wolframalpha.com/input/?i=plot+x+lnx+%2B+2x+ln+sin(1%2F(x)%5E(1%2F2)) — The Dead Legend, May 06 '17 at 10:44
@Dr.SonnhardGraubner can we apply I'hopital . Without infity÷infity — NIDHITANSH, May 06 '17 at 10:54
Good question. +1 put $x=1/t^{2}$ and the expression is $(2/t^{2})\log((\sin t)/t)$. Now use a combination of standard limits and L'Hospital's Rule to get the answer. — Paramanand Singh, May 06 '17 at 10:57

score 3 · Answer 1 · answered May 06 '17 at 11:24

3

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

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

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

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

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

answered May 06 '17 at 11:24

DonAntonio

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Thank u sir. But it also be sloved by expansions – NIDHITANSH May 06 '17 at 11:30
@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. – DonAntonio May 06 '17 at 11:33
It's very good method too. – NIDHITANSH May 06 '17 at 11:34

score 1 · Answer 2 · answered Jun 02 '20 at 03:31

You can do it without substitution, $$\lim_{x\to \infty}\left(x\ln x+2x\ln\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$ $$=\lim_{x\to \infty}2x\left(\ln\sqrt{x}+\ln\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$ $$=\lim_{x\to \infty}2x\ln\left(\sqrt{x}\sin\left(\frac{1}{\sqrt{x}}\right)\right)$$ $$=\lim_{x\to \infty}2x\ln\left(\sqrt{x}\left(\frac{(1/\sqrt{x})}{1!}-\frac{(1/\sqrt{x})^3}{3!}+\frac{(1/\sqrt{x})^5}{5!}+\ldots\right)\right)$$ $$=\lim_{x\to \infty}2x\ln\left(1-\frac{1}{6x}+\frac{1}{120x^2}-\ldots\right)$$ $$=\frac13\lim_{x\to \infty}\frac{\ln\left(1-\frac{1}{6x}\left(1+\frac{1}{20x}-\ldots\right)\right)}{\frac{1}{6x}\left(1+\frac{1}{20x}-\ldots\right)}\cdot \left(1+\frac{1}{20x}-\ldots\right)$$ $$=\frac{1}{3}(-1)\cdot 1$$$$=\color{blue}{-\frac13}$$

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

2 Answers2