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. 

$$\begin{align}\lim_{x\to 0}\frac{\sin x-x\cos x}{x\sin x}&=\lim_{x\to 0}\frac{\cos x-(\cos x-x\sin x)}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x\sin x}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x}{1+\frac{x}{\sin x}\cdot \cos x}\\&=\frac{0}{1+1\cdot 1}\end{align}$$
