# How do you use L'hospital's rule to find the limit lim_(x->0)(x-sin(x))/(x-tan(x)) ?

Apr 11, 2018

$- \frac{1}{2}$

#### Explanation:

The limit

$L = {\lim}_{x \to 0} \frac{x - \sin \left(x\right)}{x - \tan \left(x\right)}$

is of the form $\frac{0}{0}$, Thus, we can use L'hospital's rule, which says

If $f \left(a\right) = g \left(a\right) = 0$, ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)}$

Thus,

lim_(x->0)(x-sin(x))/(x-tan(x)) = lim_(x->0){d/dx(x-sin(x))}/ {d/dx(x-tan(x))}

$q \quad = {\lim}_{x \to 0} \frac{1 - \cos \left(x\right)}{1 - {\sec}^{2} \left(x\right)}$

This, again is of the $\frac{0}{0}$ form, so we use L'hospital's rule again

$L = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(1 - \cos \left(x\right)\right)}{\frac{d}{\mathrm{dx}} \left(1 - {\sec}^{2} \left(x\right)\right)} = {\lim}_{x \to 0} \frac{\sin \left(x\right)}{\left(- 2 {\sec}^{2} \left(x\right) \tan \left(x\right)\right)}$

We could use L'hospital's rule yet again, but it is much simpler to use simple trigonometry to go ahead :

L = lim_(x->0)(sin(x))/((-2sec^2(x)tan(x))) = lim_(x->0)(1)/((-2sec^3(x))

and thus

$L = - \frac{1}{2}$