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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.

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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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First, look at the two terms separately.

Then, for the second term, use the result that, for small $x$, $\sin(x) \approx x- x^3/6$.

And, of course, $\ln \sqrt{x} =\frac12 \ln x $.

After using these, recombine and see what cancels.

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  • $\begingroup$ Well, actually $x\to\infty$ ... $\endgroup$ – Michael Hoppe May 3 '15 at 16:33
  • $\begingroup$ Yes, but $1/\sqrt{x}\to 0$, so it holds. $\endgroup$ – DMcMor May 3 '15 at 16:36
  • $\begingroup$ Yup. (11 more to go) $\endgroup$ – marty cohen May 3 '15 at 16:37
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Hint:

$$ x\ln x +2x\ln \sin\frac{1}{\sqrt{x}}=x \ln \left( x \sin^2 \frac{1}{\sqrt{x}} \right)=\frac{ln \left( x \sin^2 \frac{1}{\sqrt{x}} \right)}{\frac{1}{x}}$$

Use L'H.

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We have $$\sin(1/\sqrt{x}) = \dfrac1{\sqrt{x}} - \dfrac1{3!} \dfrac1{x\sqrt{x}} + \mathcal{O}\left(1/x^{5/2} \right)$$ We have \begin{align} \ln(\sin(1/\sqrt{x})) & = \ln\left(\dfrac1{\sqrt{x}} - \dfrac1{3!} \dfrac1{x\sqrt{x}} + \mathcal{O}\left(1/x^{5/2} \right)\right)\\ & = \ln\left(1/\sqrt{x} \right) + \ln\left(1-1/(6x) + \mathcal{O}(1/x^2)\right) = -\dfrac{\ln(x)}2 - \dfrac1{6x} + \mathcal{O}(1/x^2) \end{align} Hence, we have $$x\ln(x)+2x\ln(\sin(1/\sqrt{x})) = x\ln(x) - x\ln(x) - \dfrac13 + \mathcal{O}(1/x) = - \dfrac13 + \mathcal{O}(1/x)$$ Hence, as $x \to \infty$, the limit is $-\dfrac13$.

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