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The value of $$\lim_{y\rightarrow \infty}y \ln\bigg(\frac{\sin(x+1/y)}{\sin x}\bigg), x\in \bigg(0,\frac{\pi}{2}\bigg)$$

Try: Put $\displaystyle y=\frac{1}{z}$

So $$\lim_{z\rightarrow 0}\ln\bigg(\frac{\sin(x+z)}{\sin x}\bigg)\cdot \frac{1}{z}$$

$$ = \lim_{z\rightarrow 0}\frac{\ln(\sin (x+z))-\ln(\sin x)}{z}$$

Using L'Hospital rule

$$\lim_{z\rightarrow 0}\frac{\cos(x+z)}{\sin (x+z)}=\cot x$$

Could some help me? How can I solve without L'Hospital rule? Thanks

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$\displaystyle\lim_{y\rightarrow \infty}y \ln\bigg(\frac{\sin(x+1/y)}{\sin x}\bigg)=\displaystyle\lim_{y\rightarrow \infty}y \ln\bigg(\frac{\sin(x)\cos(1/y)+\cos(x)\sin(1/y)}{\sin x}\bigg)\\=\displaystyle\lim_{y\rightarrow \infty}y \ln\bigg(\cos(1/y)+\cot(x)\sin(1/y)\bigg)\\=\displaystyle\lim_{y\rightarrow \infty}y \ln\bigg(1+\big[\cos(1/y)+\cot(x)\sin(1/y)-1\big]\bigg)$

Since $\cos(1/y)+\cot(x)\sin(1/y)-1\to0$ as $y\to\infty$, using the standard limit $\displaystyle\lim_{m\to0}\frac{\ln(1+m)}{m}=1$, we get

$=\displaystyle\lim_{y\rightarrow \infty}y\big(\cos(1/y)+\cot(x)\sin(1/y)-1\big)\\=\displaystyle\lim_{z\to0^+}\frac{\cos z-1}z+\cot(x)\cdot\lim_{z\to0^+}\frac{\sin z}z$

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  • $\begingroup$ This is how I would have answered but I guess I was late. +1 $\endgroup$ – Paramanand Singh Jan 5 at 19:37
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$\displaystyle\lim_{z\rightarrow 0}\frac{\ln(\sin (x+z))-\ln(\sin x)}{z}$ is precisely the derivative of $\ln(\sin x)$ for $x\in\left(0,\dfrac{\pi}{2}\right)$using limits.

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    $\begingroup$ Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do. $\endgroup$ – mathcounterexamples.net Jan 5 at 9:26
  • $\begingroup$ @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage. $\endgroup$ – Yadati Kiran Jan 5 at 9:36
  • $\begingroup$ @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well. $\endgroup$ – Paramanand Singh Jan 5 at 18:58
  • $\begingroup$ You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer. $\endgroup$ – Paramanand Singh Jan 5 at 19:00
  • $\begingroup$ @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate. $\endgroup$ – Yadati Kiran Jan 5 at 19:32
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Beside the simple solution given by Yadati Kiran, you could use Taylor series $$\frac{\sin(x+z)}{\sin (x)}=\cot (x) \sin (z)+\cos (z)$$ Now, using the classical expansion of $\sin(z)$ and $\cos(z)$ you then have $$\cot (x) \sin (z)+\cos (z)=1+z \cot (x)-\frac{z^2}{2}+O\left(z^3\right)$$ Continue with the logarithm to get $$\log (\cot (x) \sin (z)+\cos (z))=z \cot (x)-\frac{1}{2} z^2 \left(\cot ^2(x)+1\right)+O\left(z^3\right)$$ $$\frac 1 z \log (\cot (x) \sin (z)+\cos (z))= \cot (x)-\frac{1}{2} z \left(\cot ^2(x)+1\right)+O\left(z^2\right)$$ which shows the limit and also how it is approached.

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This is not a solution, more of a question.

I get to

1)$\log \left ( \cos (1/y) +\frac{\sin (1/y)}{1/y}\frac{\cot x}{y}\right )^y.$

$y \rightarrow \infty : $

Known:

2) $\lim_{y \rightarrow \infty} \log (1+(\cot x)/y)^y= \cot x$.

(Recall $\lim_{y \rightarrow \infty}(1+a/y)^y= e^a$, $a$ real.)

Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?

Could not come up with a solution.

Thanks

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  • $\begingroup$ The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) \to 0$ as $y\to\infty $ then $\{f(y)\} ^y\to 1$ as $y\to \infty$ (prove this!). Use $$f(y) =\dfrac{\cos(1/y)+\sin(1/y)\cot x} {1+\dfrac{\cot x} {y}} $$ and show that $y(f(y) - 1)\to 0$ as $y\to\infty $. $\endgroup$ – Paramanand Singh Jan 5 at 19:23
  • $\begingroup$ Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit . $\endgroup$ – Peter Szilas Jan 5 at 19:55

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