Easier way to solve $\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$ using L'Hospital's Rule? This is what I did for:
$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$$


*

*Check form: $\infty - \infty$.

*Rearrange it to be a quotient: 
$$\begin{align} \\
& =\lim \limits_{x \to 0} \space {\cos{x} \over \sin{x}}-{1 \over x} \\
&\\
& =\lim \limits_{x \to 0} \space {x\cos{x} - \sin{x} \over x\sin{x}} \\
\end{align}$$

*Check form: $0 \over 0$

*Apply L'Hospital's Rule:
$$ = \lim \limits_{x \to 0} \space {(-x\sin{x} + \cos{x}) - \cos{x} \over (x\cos{x} + \sin{x})} $$

*Check form: $0 \over 0$.

*Apply L'Hospital's Rule Again:
$$ = \lim \limits_{x \to 0} \space {((-x\cos{x} -\sin{x})  +\sin{x}) -\sin{x} \over (-x\sin{x} + \cos{x}) + \cos{x}} $$

*Check form: $0\over2$.


Therefore:
$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x} = 0$$
This works, but seems that apply the rule twice made things really messy. Was there a simplification step that I missed along that way that would have made this easier to deal with (other than removing the parenthesis that I put in while using the power rule)?

Speaking of simplicity, often by L'Hospital rule can be applied in combination with so-called remarkable limits. If this exercise can do so:
$$\lim \limits_{x \to 0}( \cot{x}-{1 \over x})=\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x\sin x} = \lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^2}\cdot\lim \limits_{x \to 0}\frac{x}{\sin x}=$$$$=\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^2} = \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{2x}= \lim \limits_{x \to 0}\frac{-x\sin x}{2x}=0.$$
Simplification is more obvious in the case of the more complicated example:
$$\lim \limits_{x \to 0}( \cot^2 x-{1 \over x^2}) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} = \lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} \cdot\lim \limits_{x \to 0}\frac{x^2}{\sin^2 x}=\lim \limits_{x \to 0} (\cos x +\frac{\sin x}{x})\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} = 2 \cdot \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{3x^2}= $$$$=-\frac{2}{3} $$
