Calculate $\lim_{x\to 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$ without Taylor's theorem or L'Hospital rule 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
 A: 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.
A: 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
$$
f(x)=\frac{4}{\pi}\arctan\left(\frac{\arctan x}{x}\right)-1,\qquad a=0.
$$
