Calculating $\lim_{x\to 0} \left[\frac{1}{\sin^2x}-\frac{1}{x^2}\right]$

Question

How can I calculate this limit?
$$\lim_{x\to 0} \left[\frac{1}{\sin^2x}-\frac{1}{x^2}\right] = \,?$$

I don't have any idea how to do it.

Answer 1

$$\lim_{x\rightarrow0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=\lim_{x\rightarrow0}\left(\frac{x-\sin{x}}{x^3}\cdot\frac{1+\frac{\sin{x}}{x}}{\frac{\sin^2x}{x^2}}\right)=$$ $$=2\lim_{x\rightarrow0}\frac{1-\cos{x}}{3x^2}=\lim_{x\rightarrow0}\frac{\sin^2\frac{x}{2}}{3\cdot\frac{x^2}{4}}=\frac{1}{3}.$$

Answer 2

The numerator is $$(x+\sin (x))(x-\sin (x))\sim 2x\cdot\frac {x^3}{6} $$

the denominator is $$x^2\sin^2 (x)\sim x^4$$

the result is $ \frac {1}{3}. $

Answer 3

Note that

$$\sin^2 x=\left(x-\frac{x^3}6+o(x^3)\right)^2=x^2-\frac{x^4}{3}+o(x^4)$$

$$\frac{1}{\sin^2(x)}-\frac{1}{x^2}=\frac{x^2-\sin x^2}{x^2\sin x^2}=\frac{x^2-x^2+\frac{x^4}{3}+o(x^4)}{x^4+o(x^4)}=\frac{\frac{1}{3}+o(1)}{1+o(1)}\to \frac13$$