# Thread: Trigo- limit( tan and sin)

1. ## Trigo- limit( tan and sin)

Calculate :

$\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)$

2. Originally Posted by dhiab
Calculate :

$\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)$
use L'Hopital's rule ... more than once in this case.

you should get the limit to be -2

3. Originally Posted by dhiab
Calculate :

$\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)$
If you evaluate it, you end up with the indeterminate case $\displaystyle \frac{0}{0}$.

Now, you can apply L'Hôpital's rule to get

$\displaystyle \lim_{x\to0}\frac{1-\sec^2x}{1-\cos x}=\lim_{x\to0}\frac{-\left(\sec^2x-1\right)}{1-\cos x}=\lim_{x\to0}\frac{-(\sec x+1)(\sec x-1)}{1-\cos x}$ $\displaystyle =\lim_{x\to0}\frac{-(\sec x+1)(1-\cos x)}{\cos x(1-\cos x)}=\lim_{x\to0}-\frac{\sec x+1}{\cos x}=-2$

Does this make sense?