Limit without L'Hospital rule

Question

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

Answer 1

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

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

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

Answer 2

$\displaystyle\lim_{z\rightarrow 0}\frac{\ln(\sin (x+z))-\ln(\sin x)}{z}$ is precisely the derivative of $\ln(\sin x)$ for $x\in\left(0,\dfrac{\pi}{2}\right)$using limits.

Answer 3

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.

Answer 4

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