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Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$

My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}=\lim\limits_{x\to 0} \dfrac{\frac{1}{\sqrt{1-x^2}}-1}{5x^4}$ .
Is my approach correct?

Martin Sleziak
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PrimoRaj
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    What's your reasoning behind the first step? I'm pretty sure it's wrong. Also, have you learned Taylor series yet? – JimmyK4542 Dec 15 '19 at 02:50
  • @JimmyK4542 $\lim\limits_{x\to 0} \dfrac{\sin x}{x}=1$ – PrimoRaj Dec 15 '19 at 02:53
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    You cannot factor out a $\sin x$ in the numerator so your reasoning is wrong. – Toby Mak Dec 15 '19 at 03:07
  • @TobyMak Can you give me an example where it is false? – PrimoRaj Dec 15 '19 at 03:26
  • Well, your second step is already false. If you want to write another solution, I think the answers here have covered the topic really well. Using series is the simplest and fastest method, unless you want to apply L'Hopital's rule six times. – Toby Mak Dec 15 '19 at 03:28
  • @TobyMak Can not we write $\lim\limits_{x\to 0} f(x)g(x)=\lim\limits_{x\to 0} \dfrac{f(x)}{h(x)}\lim\limits_{x\to 0}g(x)h(x)$, if we assume all the limits exist? – PrimoRaj Dec 15 '19 at 03:34
  • @PrimoRaj Yes, you can. – Toby Mak Dec 15 '19 at 03:35
  • @TobyMak I dit the same here, $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}$. – PrimoRaj Dec 15 '19 at 03:39
  • You should ask another question since this is not related to your original question. – Toby Mak Dec 15 '19 at 03:41
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    An example where replacing $\sin(x)/x$ by $1$ in a similar, but simpler expression changes the limit: $\frac{\frac{\sin(x)}{x}(1+x^4)-1}{x^2}\to -1/6$ as $x\to0$, but $\frac{1\cdot (1+x^4)-1}{x^2}\to0$ as $x\to0$. – egorovik Dec 15 '19 at 04:17
  • Using $x=\sin t$ reduces it to a famous problem from Hardy's A Course of Pure Mathematics. See https://math.stackexchange.com/q/437926/72031 – Paramanand Singh Dec 16 '19 at 02:27

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Note,

$$\sin x\sin^{-1}x=(x-\frac{x^3}6 +\frac{x^5}{120}+O(x^7)) (x+\frac{x^3}6 +\frac{3x^5}{40}+O(x^7))=x^2+\frac{x^6}{18}+O(x^8)$$

Thus,

$$\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6} =\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{x^6}{18}+O(x^8)\right)=\dfrac{1}{18}$$

Quanto
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  • Is it possible to evaluate by any other method without using ordered notation? @ Quanto – PrimoRaj Dec 15 '19 at 03:55
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    @PrimoRaj No doubt you could do it with enough applications of L'Hôpital's rule, but why? Big-O notation is by far the easiest way of reasoning with error terms for this sort of problem. – Steven Stadnicki Dec 15 '19 at 04:07
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When $x$ is small, $\sin x \approx x-x^3/6+x^5/120$ and $\sin^{-1} x\approx x+x^3/6+3x^5/40$ Then $$\lim_{x \rightarrow 0} \frac{(x-x^3/6+x^5/120)(x+x^3/6+3x^5/40)-x^2}{x^{6}}= \lim_{x\rightarrow 0}\frac{x^6/18+(.)x^8}{x^6}=\frac{1}{18}.$$

Z Ahmed
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The problem is not too difficult if you compose Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$ $$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\frac{5 x^7}{112}+O\left(x^9\right)$$ $$\sin(x)\sin ^{-1}(x)=x^2+\frac{x^6}{18}+\frac{x^8}{30}+O\left(x^{10}\right)$$ will give not only the limit but also how it is approached.

Claude Leibovici
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