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Trigonometric Limits

Posted December 5th, 2009 by alpha
in

Find:
$ \mathop{\lim}\limits_{x \to \frac{\pi}{6}}\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx} $

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answer

On December 6th, 2009 Structure says:
$$ \lim_{x \to \frac{\pi}{6}}\frac{\sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-\cosx}=\lim_{x \to \frac{\pi}{6}}\frac{\sin(x-\frac{\pi}{6})}{\cos \frac{\pi}{6}-\cosx}=<br /> \lim_ {x \to \frac{\pi}{6}}\frac{2\sin(\frac{x}{2}-\frac{\pi}{12})\cos(\frac{x}{2}-\frac{\pi}{12})}{2\sin(\frac{x}{2}-\frac{\pi}{12})\sin(\frac{x}{2}+\frac{\pi}{12})}=\frac{1}{\sin\frac{\pi}{6}}=2$$
$$\lim_{x\to a}\frac{\sin (x-a)}{\cos a-\cos x}=\lim_{x\to a}\frac{\frac{\sin (x-a)}{x-a}}{\frac{\cos a-\cos x}{x-a}}=\frac{\lim_{x\to a}\frac{\sin (x-a)}{x-a}}<br /> {\lim_{x\to a}\frac{2\sin\frac{x-a}{2}\sin\frac{x+a}{2}}{x-a}}=\frac{1}{\sin a}$$

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