limits problem

**stud_02** · May 11th 2009, 04:22 PM

$ \lim_{x \to \pi/3}\frac{\tan^3(x) -3\tan(x)}{\cos(x+\pi/6)} $

help please

**Banned for attempted hacking** · May 11th 2009, 04:29 PM

Take the derievative of the top and bottom.

In other words Apply L' Hospitals Rule, it becomes and elementary problem.

$ 3\lim_{x \to \pi/3}\frac{\tan^{2}(x)sec^{2} -3\sec^{2}x}{\-sin(x+\pi/6)} $

Evaluate it and ta-da.

The answer is negative 12 i believe but don't trust me, do it yourselve.

**galactus** · May 11th 2009, 04:38 PM

$\lim_{x\to \frac{\pi}{3}}\frac{tan^{3}(x)-3tan(x)}{cos(x+\frac{\pi}{6})}$

Have you learned L'Hopital's rule yet?.

Differentiate top and bottom:

$\lim_{x\to \frac{\pi}{3}}\frac{-3(tan^{4}(x)-1)}{sin(x+\frac{\pi}{6})}$

Now, just plug in x=Pi/3 and you got it. You should get -24

**Banned for attempted hacking** · May 11th 2009, 04:47 PM

-24 is correct. told you not to trust me.

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