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TDHawkes · Jul 15, 2007, 06:57 PM

I need to find the limit as x approaches 0 for tan 6x / sin 2x

s_cianci · Jul 15, 2007, 07:37 PM

Try L'Hopital's Rule.

galactus · Jul 15, 2007, 08:05 PM

Here's a way without L'Hopital.

$\lim_{x\to\0}\frac{tan(6x)}{sin(2x)}$

$\lim_{x\to\0}\frac{sin(6x)}{cos(6x)}\cdot\frac{1}{ sin(2x)}$

$\lim_{x\to\0}\frac{sin(6x)}{sin(2x)}\cdot\frac{1}{ cos(6x)}$

$\lim_{x\to\0}\frac{\frac{sin(6x)}{x}}{\frac{sin(2x )}{x}}\cdot\lim_{x\to\0}\frac{1}{cos(6x)}$

Since $\lim_{x\to\0}\frac{sin(x)}{x}=1$ :

$\lim_{x\to\0}\frac{6\cdot\frac{sin(6x)}{6x}}{2\tim \frac{sin(2x)}{2x}}\cdot\lim_{x\to\0}\frac{1}{cos( 6x)}=\frac{6\cdot{1}}{5\cdot{1}}\cdot{1}=\fbox{3}$