Limit without L'Hospital rule

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

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

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

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