# Calculate $\lim_{x\to 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$ without Taylor's theorem or L'Hospital rule [closed]

Calculate this limit without using taylor or hopital

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, iadvd, Claude Leibovici, Parcly Taxel, R_D
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 – Claude Leibovici Oct 2 '16 at 13:31
• $\frac{4}{\pi}\arctan\left(\frac{\arctan x}{x}\right)$ is an even function, hence if the limit exists, it must be zero. – Jack D'Aurizio Oct 2 '16 at 13:48
• 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. – Carl Mummert Oct 2 '16 at 13:54

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.
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 $$f(x)=\frac{4}{\pi}\arctan\left(\frac{\arctan x}{x}\right)-1,\qquad a=0.$$