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

Question

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 1

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

Answer 2

$$\frac{\cos x-1}{\sin^2 x}=\frac{\cos x-1}{x^2}\frac{x^2}{\sin^2 x}$$

Both $\frac{1-\cos x}{x^2}$ and $\frac{\sin x}{x}$ are basic limits.

Answer 3

Use:

• $$\frac{\cos(x)-1}{\sin^2(x)}=-\frac{\sec^2\left(\frac{x}{2}\right)}{2}$$
• $$\arctan(-x)=-\arctan(x)$$

So, we get:

$$\arctan\left(2\cdot\frac{\cos(x)-1}{\sin^2(x)}\right)=\arctan\left(-\sec^2\left(\frac{x}{2}\right)\right)=-\arctan\left(\sec^2\left(\frac{x}{2}\right)\right)$$

Answer 4

Hint : $$\lim_{x\to 0}\frac{\sin^2x}{x^2}=1$$ $$\lim_{x\to 0}\frac{1-\cos x }{x^2}=\frac12$$

Answer 5

Hint:

$$\dfrac{\cos x-1}{\sin^2x}=-\dfrac1{1+\cos x}$$ for $\cos x-1\ne0$

and

$$\arctan(-x)=-\arctan x$$

Answer 6

$(\cos x - 1) / \sin ^2x $

$= (\cos x- 1) / (1 - \cos^2x)$

$ = (\cos x - 1) / [(1 - \cos )(1 + \cos )]$

$ = - 1 / (\cos x + 1)$

Answer 7

Note that $\cos(x)-1=-2\sin^2(x/2)$. Therefore, we have

$$\begin{align} \frac{\cos(x)-1}{\sin^2(x)}&=-\frac{2\sin^2(x/2)}{\sin^2(x)}\\\\ &=-\frac12 \left(\left(\frac{\sin(x/2)}{(x/2)}\right)\,\left(\frac{x}{\sin(x)}\right)\right)^2 \end{align}$$