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Calculate this limit without using taylor or hopital

$$\lim_{x\rightarrow 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$$

I have no idea to start the problem please help

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closed as off-topic by Carl Mummert, iadvd, Claude Leibovici, Parcly Taxel, R_D Oct 4 '16 at 9:22

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    $\begingroup$ What are the methods you are supposed to know ? As commented earlier in a previous post of yours, good luck without using taylor or hopital. By the way, the names are Taylor and L'Hospital or L'Hôpital $\endgroup$ – Claude Leibovici Oct 2 '16 at 13:31
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    $\begingroup$ $\frac{4}{\pi}\arctan\left(\frac{\arctan x}{x}\right)$ is an even function, hence if the limit exists, it must be zero. $\endgroup$ – Jack D'Aurizio Oct 2 '16 at 13:48
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    $\begingroup$ Please edit the post to include more context. Even if you cannot solve it, you can explain where you encountered it and why it is of interest. If it is just a homework problem, you should know that this is a site for people to ask questions about math that they are engaged in, but it is not a homework help site. Posts that do nothing but state a problem are discouraged. $\endgroup$ – Carl Mummert Oct 2 '16 at 13:54
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We can proceed as follows \begin{align} L &= \lim_{x \to 0^{+}}\dfrac{\dfrac{4}{\pi}\arctan\left(\dfrac{\arctan x}{x}\right) - 1}{x}\notag\\ &= \lim_{x \to 0^{+}}\frac{4}{\pi}\cdot\dfrac{\arctan\left(\dfrac{\arctan x}{x}\right) - \arctan 1}{x}\notag\\ &= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\arctan\left(\frac{\arctan x - x}{\arctan x + x}\right)\tag{1}\\ &= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\cdot\dfrac{\arctan x - x}{\arctan x + x}\cdot\dfrac{\arctan\left(\dfrac{\arctan x - x}{\arctan x + x}\right)}{\dfrac{\arctan x - x}{\arctan x + x}}\tag{2}\\ &= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\cdot\dfrac{\arctan x - x}{\arctan x + x}\tag{3}\\ &= \frac{4}{\pi}\lim_{x \to 0^{+}}\dfrac{\arctan x - x}{x^{2}}\cdot\frac{x}{\arctan x + x}\notag\\ &= \frac{4}{\pi}\lim_{x \to 0^{+}}\dfrac{\arctan x - x}{x^{2}}\cdot\dfrac{1}{\dfrac{\arctan x}{x} + 1}\notag\\ &= \frac{4}{\pi}\cdot 0 \cdot\frac{1}{1 + 1}\notag\\ &= 0\notag \end{align} We have made use of the standard limit $$\lim_{x \to 0}\frac{\arctan x}{x} = 1$$ and also note that from this answer we have $$\lim_{x \to 0^{+}}\frac{\arctan x - x}{x^{2}} = 0$$ and hence $$\lim_{x \to 0^{+}}\frac{\arctan x - x}{\arctan x + x} = \lim_{x \to 0^{+}}\frac{\arctan x - x}{x^{2}}\cdot x\cdot\dfrac{1}{\dfrac{\arctan x}{x} + 1} = 0$$ and therefore the steps from $(1)$ to $(2)$ to $(3)$ are justified.

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Hint. For any differentiable function $f$ near $a$, one has $$ \lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}=f'(a). $$ One may just apply it with $$ f(x)=\frac{4}{\pi}\arctan\left(\frac{\arctan x}{x}\right)-1,\qquad a=0. $$

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  • $\begingroup$ The right question is study the differentiability !! $\endgroup$ – user315918 Oct 2 '16 at 14:04
  • $\begingroup$ @user315918 OK, I see ;) But why then refuse a Taylor series expansion? $\endgroup$ – Olivier Oloa Oct 2 '16 at 14:06
  • $\begingroup$ Because it's refused by teachers $\endgroup$ – user315918 Oct 2 '16 at 14:37

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