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How to evaluate the following limit without L'Hôpital's rule ?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$

My attempt: $$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\left(\frac{x^2}{(\sin^{-1}(x))^2x^2} - \frac1{x^2}\right)\tag2\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot \frac1{x^2} - \frac1{x^2}\tag3\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag4\\& =\left(1\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag5\\& =\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2} \tag6\\& =0\tag7\end{align}$$

But later I realized, $\lim\limits_{x\to a} f(x)\ g(x) = \lim\limits_{x\to a} f(x)\lim\limits_{x\to a} g(x) $ if and only if both $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{x\to a} g(x)$ are well defined and finite. Thus moving from $(3)$ to $(4)$ is absolutely wrong.

I don't have any more ideas about the problem.

Utkarsh
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3 Answers3

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$$\lim_{x \rightarrow 0} \left( \frac{1}{\arcsin(x)^2} - \frac{1}{x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{x^2 - \arcsin(x)^2}{\arcsin(x)^2 x^2} \right)$$ Taylor series of $ \arcsin(x) = x + x^3/6 + O(x^5) $ at $x \rightarrow 0$, in this way $$\lim_{x \rightarrow 0} \left( \frac{x^2 - x^2 - x^4/3 - x^6/36}{(x + x^3/6)^2x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3 - x^6/36}{x^4 + x^6/3 + x^8/36} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3}{x^4} \right) = - \cfrac{1}{3}$$

Gary
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Matvei
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    I belive Taylor series are stronger than L'Hôpital's rule, and the OP is not even allowed to use L'Hôpital's rule. – Gary Oct 26 '22 at 02:56
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Using $$ \sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3 \, x^5}{40} + \mathcal{O}(x^7) $$ then $$\frac{1}{(\sin^{-1}(x))^2} = \frac{1}{x^2} - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6). $$ This gives $$ \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} = - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6) $$ and the limit $$ \lim_{x \to 0} \left( \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} \right) = - \frac{1}{3}.$$

Leucippus
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    I belive Taylor series are stronger than L'Hôpital's rule, and the OP is not even allowed to use L'Hôpital's rule. – Gary Oct 26 '22 at 03:14
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Let $\arcsin x=t$ and then $x=\sin t$. So \begin{eqnarray} &&\lim_{x \rightarrow 0} \left( \frac{1}{\arcsin(x)^2} - \frac{1}{x^2} \right) \\ & =& \lim_{t \rightarrow 0}\frac{\sin^2t - t^2}{t^2 \sin^2t}\\ & =& \lim_{t \rightarrow 0}\frac{\sin t - t}{t^3}\cdot\frac{\sin t+t}{t}\cdot\frac{t^2}{ \sin^2t}\\ &=&-\frac16\cdot2=-\frac13. \end{eqnarray} Here $$ \lim_{t \rightarrow 0}\frac{\sin t - t}{t^3}=-\frac16 $$ is used from here or here.

xpaul
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