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I am trying to find this limit,

$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$

Using the fundamental theorem of calculus, part 1, $\arctan$ is a continuous function, so $$F(x):=\int_0^x \arctan{t}dt$$ and I can change the limit to $$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$

I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.

I tried using l'Hospital's rule, but the calculation gets tedious.

Can anyone give me hints?

EDIT

I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!

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  • $\begingroup$ Numerical evaluation says the limit should be $1/6$. $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 23:34
  • $\begingroup$ @SimplyBeautifulArt How did you do it? What kind of method did you use for numerical evaluation? $\endgroup$ – zxcvber Mar 29 '17 at 23:36
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    $\begingroup$ Calculators of course. $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 23:37
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Power series can help, if you know them.

We know that $$ \arctan t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+1}}{2n+1},\qquad \lvert t\rvert<1. $$ Therefore an antiderivative for $\arctan t$ is $$ F(t):=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+2}}{(2n+1)(2n+2)},\qquad \lvert t\rvert<1. $$ So, as $x\to0$, $$ F(x)=\frac{x^2}{2}-\frac{x^4}{12}+O(x^6). $$ Now, recalling that $$ \sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+O(x^5)=x+O(x^3), $$ we see that as $x\to0$ (and therefore $\sin x\to0$), $$ \begin{align*} F(\sin x)&=\frac{1}{2}\left(x-\frac{x^3}{6}+O(x^5)\right)^2-\frac{1}{12}\left(x+O(x^3)\right)^4+O[(x+O(x^3))^6]\\ &=\frac{1}{2}\left[x^2-\frac{x^4}{3}+O(x^6)\right]-\frac{1}{12}\left[x^4+O(x^6)\right]+O(x^6)\\ &=\frac{x^2}{2}-\frac{x^4}{4}+O(x^6) \end{align*} $$ So, all told, our limit is $$ \lim_{x\to0}\frac{F(x)-F(\sin x)}{x^4}=\lim_{x\to0}\frac{\,\frac{x^4}{6}\,}{x^4}=\frac{1}{6} $$

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  • $\begingroup$ I really like the way you used $O(x^n)$ notation here. Thanks! $\endgroup$ – zxcvber Mar 29 '17 at 23:56
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Here's one way to do this. The Taylor expansion for $\arctan(x) = x - x^3/3 + x^5/5 + \cdots$. So then $$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12 + x^6/30) - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{12} - \frac{(\sin x)^6}{30}.$$

Using another Taylor expansion, $\sin x = x - x^3/3! + x^5/5! + \cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$), $$ (\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$ and $$ (\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$ Thus $$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$

So multiplying by $x^{-4}$ and taking the limit gives $1/6$.

As an aside, l'Hopital's rule would also work.

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Since $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get

$$\lim_{x \to 0}\frac{F(x)-F(\sin x)}{x^4} = \lim_{x \to 0}\frac{\arctan x - \cos x\arctan \sin x}{4x^3}.$$

This last limit can be evaluated using Taylor series:

$$\arctan x = x-\frac{x^3}{3}+O(x^5) $$ and

$$\cos x \arctan \sin x = x-x^3+O(x^5) $$

and the limit you are looking for is equal to

$$\lim_{x \to 0}\frac{x-\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \frac{1}{6}. $$

The ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\sin$, $\cos$ and $\arctan$. Easier done than said.

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  • $\begingroup$ Actually, you just take the derivatives directly instead of trying to combine well-known expansions since that's ugly and we only need a few terms. $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 23:49
  • $\begingroup$ @SimplyBeautifulArt That's true, but I expect the third derivative of $\cos x \arctan \sin x$ to be even uglier.. $\endgroup$ – Stefano Mar 29 '17 at 23:51
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    $\begingroup$ Well, the key is that its easy to ask your calculator to take the third derivative ;) Combining Taylor expansions? Not so much. $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 23:53
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Note that by the Mean Value Theorem for integrals we have $$\int_{\sin x}^{x}\arctan t\,dt = (x - \sin x)\arctan c$$ for some $c$ between $x$ and $\sin x$. Then we have $$\lim_{x \to 0}\frac{1}{x^{4}}\int_{\sin x}^{x}\arctan t\,dt = \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{\arctan c}{x}$$ The first factor on the right tends to $1/6$ (via L'Hospital's Rule or Taylor series) and we show that next factor tends to $1$. For this we assume that $x \to 0^{+}$ so that $\sin x < c < x$ and therefore $$\arctan \sin x < \arctan c < \arctan x$$ and dividing by $x$ we get $$\frac{\arctan \sin x}{\sin x}\cdot\frac{\sin x}{x} < \frac{\arctan c}{x} < \frac{\arctan x}{x}$$ By Squeeze theorem we see that $(\arctan c)/x \to 1$ as $x \to 0^{+}$. A similar argument can be given for $x \to 0^{-}$ and we have the desired limit as $1/6$.

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