Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?

Question

Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$

My attempt:

$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$

=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)

Applying L'Hospital's Rule we get,

=$\lim_{x\to 0}(\frac{2\sin x \cos x -2x}{2x \sin^2 x+ 2x^2\sin x \cos x})$

=$\lim_{x\to 0}(\frac{\sin 2x -2}{2x \sin^2 x+ x^2\sin 2x})$ ($\frac{0}{0}$ form)

Applying L'Hospital's Rule we get,

=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+2x\sin x \cos x)+ (2x\sin 2x+2x^2\cos 2x)})$

=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ (2x\sin 2x+2x^2\cos 2x)})$

=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ 2(x\sin 2x+x^2\cos 2x)})$

=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x+x\sin 2x+x^2\cos 2x)})$

=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+x\sin 2x+ x\sin 2x+x^2\cos 2x})$

=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+2x\sin 2x+x^2\cos 2x})$ ($\frac{0}{0}$ form)

Applying L'Hospital's Rule we get,

=$\lim_{x\to 0}(\frac{-2\sin 2x}{2\sin x \cos x+2(\sin 2x+ 2x\cos 2x)+(2x\cos 2x-2x^2\sin 2x})$

=$\lim_{x\to 0}(\frac{-2\sin 2x}{3\sin 2x+6x\cos 2x-2x^2\sin 2x})$ ($\frac{0}{0}$ form)

Applying L'Hospital's Rule we get,

=$\lim_{x\to 0}(\frac{-4\cos 2x}{6\cos 2x+6(\cos 2x-2x\sin 2x)-2(2x\sin 2x+2x^2\cos 2x)})$

=$\lim_{x\to 0}(\frac{-4\cos 2x}{12\cos 2x-12x\sin 2x-4x\sin 2x-4x^2\cos 2x)})$

=$\frac{-4\cos 0}{12 \cos 0-0-0-0}$

=$\frac{-4}{12}$

=$\frac{-1}{3}$

My problem: This method is very lengthy and involves a lot of calculations. Is there is any other (better and efficient) method to evaluate this limit?

Answer 1

$$\lim_{x\to 0}\frac{\sin x-x}{x^3}=\lim_{x\to 0}\frac{\cos x-1}{3x^2}=\lim_{x\to 0}\frac{-\sin x}{6x}=-\frac{1}{6}$$

\begin{align*} \lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)&=\lim_{x\to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}\\ &=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x\sin x+x^2}{\sin^2x}\right)\\ &=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x}{\sin x}+\lim_{x\to 0}\frac{x^2}{\sin^2x}\right)\\ &=\left(-\frac{1}{6}\right)(1+1)\\ &=-\frac{1}{3} \end{align*}

Answer 2

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

You have that $\sin(x)=x-\frac{x^3}{3!}+o(x^3)$, and thus $$\sin^2(x)=x^2-\frac{x^4}{3}+o(x^4).$$ Finally, $$\frac{1}{x^2}-\frac{1}{\sin^2(x)}=\frac{-x^4+o(x^4)}{3x^2\sin^2(x)}\underset{x\to 0}{\longrightarrow }-\frac{1}{3}$$

Answer 3

$$=\left(\frac{x}{\sin x}\right)^2\left(\frac{\sin x}{x}+1\right)\left(\frac{\sin x -x}{x^3}\right)\to 1^2(1+1)\left(-\frac{1}{6}\right)=-\frac{1}{3}$$