# Calculating $\lim\limits_{x\to\pi/6^+}\frac{|1-2\sin x|}{4\cos^2 x-3}$ [closed]

I'm finding this limit problem confusing.

$$\lim_{x\to\pi/6^+}\frac{|1-2\sin x|}{4\cos^2 x-3}.$$

The answer is $\dfrac{-1}{2}$.

I tried the $1 - \sin^2x$ but I keep getting $1$ as my answer. What I did was cancel out the $\sin x$ with $\sin^2x$ then plugged in the limit. I'm doing it wrong so I need more help.

Hint:$$\lim_{x\rightarrow \frac{\pi^+}{6}}\frac{2\sin x-1}{1-4\sin^2x}$$ $$\lim_{x\rightarrow \frac{\pi}{6}}\frac{2\cos x}{-4\sin(2x)}\tag{Using L'Hospital}$$
$$RHL=\frac{-1}{2}$$
Edit: Without L'hospital, factorize and proceed $$\lim_{x\rightarrow \frac{\pi}{6}}\frac{2\sin x-1}{(1-2\sin x)(1+2\sin x)}$$
• It flew into ashes. Note: $4 \cos^2 x -3=1-4\sin^2x$@j.doe Apr 29 '17 at 4:00
• As i took Right hand limit, at that time $2 \sin x>1$ thus we reached the condition where $|\alpha|$ where $\alpha <0$ thus we place $-\alpha$.@j.doe Apr 29 '17 at 4:24