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$$\lim_{x\to0}\frac{\arcsin x-\sin x}{x^3}$$ without using series or L'Hospital

Is there any ohter simpler method? Expansion of $\arcsin$ is not trivial like tha of sine and L'Hospital is too cumbersome here.

Source-Question $2.10$

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  • $\begingroup$ You will only need to apply L'Hospital twice $\endgroup$ – Shubham Johri Feb 11 at 17:54
  • $\begingroup$ @ShubhamJohri Doesn't the whole thing look ugly? $\endgroup$ – tatan Feb 11 at 17:56
  • $\begingroup$ It's as easy with L'Hospital as any answer below $\endgroup$ – Shubham Johri Feb 11 at 18:01
  • $\begingroup$ @ShubhamJohri Maybe. But the answer below is more elegant. $\endgroup$ – tatan Feb 11 at 18:07
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    $\begingroup$ See math.stackexchange.com/questions/387333/… $\endgroup$ – lab bhattacharjee Feb 11 at 18:25
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This site has repeatedly shown, without the methods forbidden in this question, that $\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}$. Hence $$\lim_{x\to 0}\frac{\arcsin x-x}{x^3}=\lim_{y\to 0}\frac{y-\sin y}{\sin^3 y}=\lim_{y\to 0}\frac{y-\sin y}{y^3}\left(\frac{y}{\sin y}\right)^3=\frac{1^3}{6}=\frac{1}{6}.$$Summing, your limit is $\frac{1}{3}$.

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    $\begingroup$ Thanks! That was clever $\endgroup$ – tatan Feb 11 at 17:53
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You will only need two repeated applications of the L'Hospital Rule.

You could substitute $\arcsin x=\alpha$ with $\alpha\to0$, you will only require the series for $\sin y$ then.

$$\lim_{\alpha\to0}\frac{\alpha-\sin\sin\alpha}{\sin^3\alpha}= \lim_{\alpha\to0}\frac{\alpha-\sin\alpha+\frac{\sin^3\alpha}{3!}...}{\sin^3\alpha}=\frac16+\lim_{y\to0}\frac{y-\sin y}{\sin^3y}=\frac16+\lim_{y\to0}\frac{y-y+\frac{y^3}{3!}...}{\sin^3y}=1/3$$

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