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:$$\lim_{x\rightarrow \frac{\pi^+}{6}}\frac{2\sin x-1}{1-4\sin^2x}$$ $$\lim_{x\rightarrow \frac{\pi}{6}}\frac{2\cos x}{-4\sin(2x)}\tag{Using L'Hospital}$$


Edit: Without L'hospital, factorize and proceed $$\lim_{x\rightarrow \frac{\pi}{6}}\frac{2\sin x-1}{(1-2\sin x)(1+2\sin x)}$$

  • $\begingroup$ where is the '-3' in the denominator ??????? $\endgroup$
    – j.doe
    Apr 29 '17 at 3:59
  • $\begingroup$ It flew into ashes. Note: $4 \cos^2 x -3=1-4\sin^2x$@j.doe $\endgroup$ Apr 29 '17 at 4:00
  • $\begingroup$ how and why did you switch around the 1 - 2sinx in the numerator to ....2sinx -1 $\endgroup$
    – j.doe
    Apr 29 '17 at 4:18
  • $\begingroup$ 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 $\endgroup$ Apr 29 '17 at 4:24
  • $\begingroup$ wait what does 'a' represent and how did it go to a<0 $\endgroup$
    – j.doe
    Apr 29 '17 at 4:35

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