calculus - Evaluate this trigonometric limit with L'hospital's rule.

I've been asked to evaluate the limit using L'hospital's rule, or any simpler more elementary method.

$$\lim\limits_{x \to 0} {{x-\sin(x)}\over{x-\tan(x)}}$$

Applying L'Hospitals rule, I get: $${1-\cos(x)}\over{1-\sec^2(x)}$$

This still yields an indeterminate form, and applying L'Hospitals rule any more times makes things get messy fast. I think I'm stuck.

Other things I have tried include writing in terms of sine and cosine, both before and after applying the rule. I can't seem to wrestle it out of an indeterminate form.

How can I proceed?

Answer

We have the following:

$$\frac{1-\cos(x)}{1-\frac{1}{\cos^{2}(x)}}=\frac{1-\cos(x)}{\frac{\cos^{2}(x)-1}{\cos^{2}(x)}}=\frac{-\cos^{2}x}{1+\cos(x)}$$

which leads to

$$\lim_{x\to 0}\frac{-\cos^{2}x}{1+\cos(x)}=\frac{-1}{2}$$

which seems to be correct, according to WolframAlpha.