$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $

Question

$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $$

My Approach:

$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{(1-\sin x)(1+\sin x)} \biggl|$$

$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{1}{\log(\sin x)}+\frac{1}{(1-\sin x)} \biggl| $$

$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{(\sin x - 1)}{(\sin x-1)\log(1+(\sin x - 1))}+\frac{1}{(1-\sin x)} \biggl|$$

I could not processed further.

By Applying L'Hôpital's Rule I get answer $0$ but given answer is $\frac{1}{4}$

Also L'Hôpital's Rule is going lengthy.

Answer 1

Let me use $\ln$ in place of $\log$ and get from your third line $$\frac{1}{\ln(\sin x)}+\frac{1}{(1-\sin x)} =\frac{1-\sin x+\ln(\sin x)}{(1-\sin x)\ln(\sin x)} = \\= \frac{1-\sin x+\sin x -1+(\sin x-1)^2\frac{1}{2}-(\sin x - 1)^3\frac{1}{3}+\cdots}{-(\sin x - 1)^2}\to -\frac{1}{2}$$

Answer 2

Let $u=\sin x \to 1$ to obtain

$$\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x}=\frac{1}{(1+u)\log u}+\frac{1}{1-u^2}=\frac{1-u+\log u}{(1-u^2)\log u} $$

and by $v=u-1 \to 0$

$$\frac{\log (1+v)-v}{-v(v+2)\log (1+v)}=\frac{\log (1+v)-v}{v^2}\frac{v}{-v(v+2)}\frac{v}{\log (1+v)}\to -\frac12\cdot -\frac12\cdot 1=\frac14$$