[Math] Easier way to solve $\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$ using L’Hospital’s Rule

Question

This is what I did for:

$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$$

1. Check form: $\infty – \infty$.
2. Rearrange it to be a quotient:
   $$\begin{align} \\
   & =\lim \limits_{x \to 0} \space {\cos{x} \over \sin{x}}-{1 \over x} \\
   &\\
   & =\lim \limits_{x \to 0} \space {x\cos{x} – \sin{x} \over x\sin{x}} \\
   \end{align}$$
3. Check form: $0 \over 0$
4. Apply L'Hospital's Rule:
   $$ = \lim \limits_{x \to 0} \space {(-x\sin{x} + \cos{x}) – \cos{x} \over (x\cos{x} + \sin{x})} $$
5. Check form: $0 \over 0$.
6. Apply L'Hospital's Rule Again:
   $$ = \lim \limits_{x \to 0} \space {((-x\cos{x} -\sin{x}) +\sin{x}) -\sin{x} \over (-x\sin{x} + \cos{x}) + \cos{x}} $$
7. 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)?

Answer 1

You did indeed miss a simplification:

$$\lim \limits_{x \to 0} \space {(-x\sin{x} + \cos{x}) - \cos{x} \over (x\cos{x} + \sin{x})}=\lim \limits_{x \to 0} \space {-x\sin{x}\over (x\cos{x} + \sin{x})}$$

It's not a lot simpler, but it is simpler.