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Why do we have$$\lim _{x\to \,0}\left(\frac{\arctan\left(x\right)-\sin\left(x\right)}{x^3}\right)=-\frac{1}{6}?$$

I used L’Hopital’s Theorem; That result is

$$\lim _{x\to 0}\left(\frac{-\frac{2x}{\left(1+x\right)^2}+\sin\left(x\right)}{6x}\right)$$

So my answer is $\frac{1}{6}$, but the answer to the result using Wolfram Alpha is $-\frac{1}{6}$.

What's wrong?

Mark Viola
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Dosung
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  • Where do you get 1/6 from? If you evaluate the limit, you should get -2/6 + 1/6 = -1/6 – Long Sep 28 '17 at 04:26
  • Write $$\arctan(x)-\sin x=\arctan(x)-x+(x-\sin x)$$ and use https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Sep 28 '17 at 06:26

2 Answers2

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HINT:

Note that we have

$$\frac{-\frac{2x}{(1+x^2)^2}+\sin(x)}{6x}=-\frac{2}{6(1+x^2)^2}+\frac16 \frac{\sin(x)}{x}$$

and

$$-\frac13+\frac16=-\frac16$$

Mark Viola
  • 179,405
2

I think the quickest way to see this is by Taylor's theorem.

By the grace of god, the first terms in the Taylor expansions cancel, making this limit finite, since near 0 $$ \arctan(x)-\sin(x)=x-x^3/3+O(x^4)-(x-x^3/6+O(x^5) $$ and so $$ \frac{\arctan(x)-\sin (x)}{x^3}\sim \frac{-x^3/3+x^3/6}{x^3}=-1/6 $$

operatorerror
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  • Thank you. but I am not familiar with Taylor's theorem. After studying more, I will refer to your answer. – Dosung Sep 28 '17 at 04:48