# Finding $\lim_{x\to 0}\Big({\frac 1{x^2} - \frac 1{\sin^2x}}\Big)$ [duplicate]

How to solve :$$\lim_{x\to0}\Big(\frac 1{x^2} - \frac 1{\sin^2x}\Big)$$ My approach : as $\lim_{x\to0}{\frac {\sin x}x}=1$. Therefore when $x\to0$ , $\sin x=x$. So when $x\to0$ , $\frac 1{X^2}=\frac 1{\sin^2x}$. So the answer to the problem is $0$. But in the book answer is given $\frac 13$ . How?

• Oops sorry I did'nt know someone already asked it. Commented Jun 7, 2015 at 8:16

You used that in $$\frac1{x^2}-\frac1{\sin^2x}=\frac1{x^2}\left(1-\frac{x^2}{\sin^2x}\right)$$ the expression parenthesis tends $\to 0$ as $x\to 0$. However, $\frac1{x^2}$ does not remain bounded and in fact grows fast enough to cancel the effect.
$$\sin x\approx x-x^3/6+...\\ \sin^2x\approx x^2(1-x^2/3+...)\\ \frac{1}{\sin^2x}\approx x^{-2}(1-x^2/3+...)^{-1}\\ =x^{-2}+1/3+...$$
• So, the limit is $-\frac 13$ which is what I was typing when arrived your answer. Can we suppose one more typo in a textbook ? Commented Jun 7, 2015 at 8:18