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?
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$\begingroup$ Oops sorry I did'nt know someone already asked it. $\endgroup$– Arpan BiswasCommented Jun 7, 2015 at 8:16
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2 Answers
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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.
answered Jun 7, 2015 at 8:14
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$\begingroup$ Thanks. That helped me understand the situation here. $\endgroup$ Commented Jun 7, 2015 at 8:18
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$$\sin x\approx x-x^3/6+...\\ \sin^2x\approx x^2(1-x^2/3+...)\\ \frac{1}{\sin^2x}\approx x^{-2}(1-x^2/3+...)^{-1}\\ =x^{-2}+1/3+...$$
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$\begingroup$ 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 ? $\endgroup$ Commented Jun 7, 2015 at 8:18
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$\begingroup$ I suppose so, I have a typo in this answer... $\endgroup$– Empy2Commented Jun 7, 2015 at 8:28