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. 

You have
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x}
$$
You can't apply L'Hopital's rules to either of the two fractions above since the numerators do not approach $0$ or $\pm\infty$.  But you can first add the fractions and then use L'Hopital's rule:
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{-\sin^2 x + x^2}{x^2\sin^2 x}
$$
