limits - Evaluating $ lim_{x to 0}{ arctan( 2 {Large[} frac{cos(x) - 1}{sin^2x} {Large]} )} $

Having
$$ \lim_{x \to 0}{ \arctan\left( 2 \left[ \frac{\cos(x) - 1}{\sin^2x}\right] \right)} = L $$

The interesting part however is $$ \frac{\cos(x) - 1}{\sin^2x} $$ and $ \lim_{x \to 0} \frac{(\cos(x) - 1)}{\sin^2x} = \left[\frac{0}{0}\right] = -1$. With l'hopital's rule it's easy to solve - I'm interested in other methods though¹. Hints are welcome too.

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Attempt 1

Considering $$ 1 = \sin^2x + \cos^2x $$ Plugging in:
$$ \frac{\cos(x) - \sin^2x - \cos^2x}{\sin^2x}
\longrightarrow \lim_{x \to 0} {\cot(x)\csc(x) - 1 - \cot^2x}$$
which doesn't go very far.

Attempt 2

Considering:

$$ -2 \leq \cos(x) - 1 \leq 0 \rightarrow \frac{-2}{\sin^2x} \leq \cos(x) - 1 \leq \frac{0}{\sin^2x}$$

But $$ \lim_{x\to0}{ \frac{-2}{\sin^2x} } \neq \lim_{x\to0}{ \frac{0}{\sin^2x}} $$

so the squeeze theorem cannot be applied. (bonus question: can it?)

^{Please avoid Taylor expansion.}

Answer

$${\cos x-1\over\sin^2x}={\cos x-1\over\sin^2x}\cdot{\cos x+1\over\cos x+1}={\cos^2x-1\over\sin^2x(\cos x+1)}={-\sin^2x\over\sin^2x(\cos x+1)}={-1\over\cos x+1}$$