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How can I find $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$?

I've tried using the Lhopital rule and it got me here:

$\lim_{x\to0}\frac{\arcsin x -x}{x^2} = \lim_{x\to0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{2x} = \lim_{x\to0}\frac{x}{2\sqrt{1-x^2}\cdot (1+\sqrt{1-x^2})}$

This doesn't make life much easier, unless I could say that $\frac{x}{2\sqrt{1-x^2}\cdot (1+\sqrt{1-x^2})}$ is continuous at $x=0$..

Is there a better way to approach this?

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  • 1
    $\begingroup$ But it is continuous at $0$. The denominator is continuous as product of continuous functions and it is not zero for $x = 0$, so we find that the limit is 0/4 = 0$. You could also use the Taylor expansion around 0 of arcsin, as user330587 shows in his answer bellow. $\endgroup$ – Student Feb 4 '17 at 10:37
  • $\begingroup$ If the limit exists it has to be zero since $\frac{\arcsin x-x}{x^2}$ is an odd function. $\endgroup$ – Jack D'Aurizio Feb 4 '17 at 19:52
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$$\arcsin(x)=x+\frac{x^3}{6}+o(x^3)$$ then $$\lim_{x\to 0}\frac{\arcsin(x)-x}{x^2}=\lim_{x\to 0}\frac{x}{6}=0.$$

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Let $y=\arcsin x$ then $$\lim_{x\to0}\frac{\arcsin x -x}{x^2}=\lim_{y\to0}\frac{y-\sin y}{\sin^2y}$$ two times L'Hopital gives us $0$.

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