# 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.

## closed as off-topic by Ken Duna, Zaid Alyafeai, Claude Leibovici, qbert, NamasteApr 29 '17 at 12:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ken Duna, Zaid Alyafeai, Claude Leibovici, qbert, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

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 – The Dead Legend 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 – The Dead Legend Apr 29 '17 at 4:24