Let $f(x) = 1/x^2 - 1/\sin ^2 x$ . Find Value of $\lim_{x \to 0} f(x)$ . I've tried L'Hôpital's rule but it didn't work . Also I was unable to use trigonometric identities.
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Use Taylor expansions. Near 0, $\sin(x)\approx x-\frac{1}{3!}x^3+O(x^5)$. – Marc Nov 23 '17 at 16:51
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What's the meaning of $O$ ? – S.H.W Nov 23 '17 at 16:53
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$O$ means terms of order $x^5$. If $x$ is small, as in this case, where $x\to0$, then the higher powers are even smaller and can be neglected. You can forget about that part and just use terms of order less than 5. Use $\sin^2(x)\approx x^2-\frac{1}{3}x^4$ and you'll find the solution. – Marc Nov 23 '17 at 16:59
2 Answers
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$$\lim_{x\to 0} (\dfrac{1}{x^2}-\dfrac{1}{\sin^2 x})=\lim_{x\to 0}\dfrac{\sin ^2x -x^2}{x^2\sin^2x}=\dfrac{-1}{3}$$
And use 4 times L'Hôpital's rule .
Or
$$\lim_{x\to 0}\dfrac{\sin ^2x -x^2}{x^2\sin^2x}=\lim_{x\to 0}\dfrac{(\sin x -x)(\sin x+x)}{x^2 \sin x}$$
$$\lim_{x\to 0}(\dfrac{(\sin x -x)}{x^3}\times\dfrac{x^3}{x^2\sin x}\times \dfrac{\sin x+x}{\sin x})=\dfrac{-1}{3}$$
that :
$$\lim_{x\to 0}(\dfrac{(\sin x -x)}{x^3}=\dfrac{-1}{6}$$
$$\lim_{x\to 0}(\dfrac{x}{\sin x})=1$$
$$\lim_{x\to 0}(\dfrac{(\sin x +x)}{\sin x}=2$$
look :
Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule
Almot1960
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\begin{align}\lim_{x\to0}\frac1{x^2}-\frac1{\sin^2x}&=\lim_{x\to0}\frac{\sin^2(x)-x^2}{x^2\sin^2x}\\&=\lim_{x\to0}\frac{\left(x-\frac{x^3}6+\cdots\right)^2-x^2}{x^2\left(x-\frac{x^3}6+\cdots\right)^2}\\&=\lim_{x\to0}\frac{-\frac{x^4}3+\cdots}{x^4+\cdots}\\&=-\frac13.\end{align}
José Carlos Santos
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