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. 

Use Taylor-MacLaurin at order $2$:
\begin{align*}\frac1x - \frac{\cos x}{\sin x}&=\frac1x - \frac{1-\dfrac{x^2}2+o(x^2)}{x-\dfrac{x^3}6+o(x^3)}=\frac1x-\frac1x\left(\frac{1-\smash[t]{\dfrac{x^2}2}+o(x^2)}{1-\smash[b]{\dfrac{x^2}6}+o(x^2)}\right)\\&=\frac1x-\frac1x\Bigl(1-\frac{x^2}3+o(x^2)\Bigr)= \frac x3+o(x)\underset{x\to 0}{\longrightarrow} 0.\end{align*}
