<s>[INST] Question on how to find this particular limit.. 
Find the following limit $$\lim_{x \to 0^-}\left(\dfrac{\cos(x)}{\sin(x)}-\dfrac{1}{x}\right)$$  

I checked online and the correct answer is $-\infty$ but I am not sure how to get to it via LHopital??  
My Attempt
As it is a $0/0$ type limit I tried using LHopital's;
$$\lim_{x \to 0^-}\left(\dfrac{\cos(x)}{\sin(x)}-\dfrac{1}{x}\right)=\left(\dfrac{x\cos(x)-\sin(x)}{x\sin(x)}\right)$$
and differentiating we get;
$$\lim_{x \to 0^-}\left(\dfrac{-x\sin(x)}{x\cos(x)+\sin(x)}\right)$$
This is still a $0/0$ type limit... differentiating further doesn't help either...  
Any help or advice would be much appreciated
 [/INST] Bully for Taylor:
$$\begin{align}
\frac{\cos x}{\sin x} - \frac{1}{x} &= \frac{x\cos x - \sin x}{x\sin x}\\
&= \frac{x\left(1-\frac{x^2}{2} + O(x^4)\right)-\left(x - \frac{x^3}{6} + O(x^5)\right)}{x\left(x - \frac{x^3}{6} + O(x^5)\right)}\\
&= \frac{-\frac{x^3}{3} + O(x^5)}{x^2\left(1-O(x^2)\right)}\\
&= -\frac{x}{3}\frac{1+O(x^2)}{1-O(x^2)},
\end{align}$$
and we find the limit for $x\to 0$ is $0$, not $-\infty$.
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