Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}=  \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks. 

$$\lim_{x\to 0} \frac{1}{x}-\frac{\cos(x)}{\sin(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{\tan(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{x+\frac{x^3}{3}+O(x^5)}=\lim_{x\to 0}\frac{1}{x}-\frac{1}{x}=0$$
