Find $ \lim\limits_{x \rightarrow 0} \frac{\cos(\sqrt{\cot^{-1}(x)})-1}{x} $ I want to find that limit
$$ \lim_{x \rightarrow 0} \frac{\cos(\sqrt{\cot^{-1}(x)})-1}{x} $$

I can use there L'Hospital's rule, so let's calculate 
$$ (\cos(\sqrt{\cot^{-1}(x)})-1)' = (\cos(\sqrt{\cot^{-1}(x)})' = \\
(\sqrt{\cot^{-1}(x)})'\cdot \sin(\sqrt{\cot^{-1}(x)}) = \\
\sin(\sqrt{\cot^{-1}(x)} \cdot \frac{1}{1+x^2}\cdot \frac{1}{2\sqrt{\cot^{-1}(x)}} $$
So
$$ \lim_{x \rightarrow 0} \frac{\cos(\sqrt{\cot^{-1}(x)})-1}{x} = \\ \lim_{x \rightarrow 0}\sin(\sqrt{\cot^{-1}(x)} \cdot \frac{1}{1+x^2}\cdot \frac{1}{2\sqrt{\cot^{-1}(x)}} = \sin(\sqrt{\pi / 2}) \cdot \frac{1}{2 \sqrt{\pi / 2}}  $$
but wolfram tells that it is $-\infty$
 A: I will assume that $\cot^{-1}$ is the inverse of the $\cot$ function.
Depending on what your definition is (more specifically the range of $\cot^{-1}$), it can be that $\cot^{-1}(x) \to \frac{\pi}{2}$ when $x \to 0$ or  $\cot^{-1}(x) \to \pm\frac{\pi}{2}$ when $x\to 0$, depending on the sign of $x$. In each case, the numerator does not tend to $0$ and you cannot use L'Hôpital's rule.
Now, if you are using the first definition, since the numerator is negative when $x \to 0$, and the denominator can be either positive or negative when $x \to 0$, the left limit and the right limit are not equal and hence the limit does not exist.
If you are using the other definition, then the under-limit function isn't defined for $x<0$ and we define the "total" limit to be the right limit, which is equal to $-\infty$.
A: It isn't an indetermined form because $\sqrt{\cot^{-1}(0)}$ isn't an angle which gives the value 1 to the cosine. So we have a finite value divided to a variable wich goes to zero, follows that the limit diverges.
