Evaluating $ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $ 
$$ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $$

I tried to do this with some kind of substitution but failed miserably. Any hints or help?
 A: Substitute $y:=\frac{1}{x+1}$ to rewrite your limit as$$\lim_{y\to0}(1-y)\frac{(\arctan 1-\arctan(1-y))}{y}=\lim_{y\to0}(1-y)\cdot\arctan^\prime1=\frac12.$$Or if we take @DinnoKoluh's approach,$$\arctan1-\arctan\frac{x}{x+1}=\arctan\frac{1}{2x+1}\approx\frac{1}{2x}.$$
A: Hint:
$$ \arctan(x)\pm\arctan(y) = \arctan(z) $$
where $z$ is:
$$ z = \frac{x\pm y}{1\mp xy} $$
And knowing the fact that:
$$ \lim_{x\to0} \frac{\arctan(f(x))}{f(x)} = 1 $$
If $ f(x) = 0 $ as $x\to0$.
This can be proven by the L'Hospital's rule:
$$ \lim_{x\to0} \frac{\arctan(f(x))}{f(x)} = \lim_{x\to0} \frac{\frac{1}{1+f^2(x)}f'(x)}{f'(x)} = \lim_{x\to0} \frac{1}{1+f^2(x)} = 1. $$
A: First set $1/x=h$
using my answer here in Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\arctan 1-\arctan\dfrac x{x+1}=\arctan1-\arctan\dfrac1{h+1}=\cdots=\arctan\dfrac h{2+h}$$
$$ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) =\lim_{h\to0^+}\dfrac{\arctan\dfrac h{2+h}}{\dfrac h{2+h}}\cdot\lim_{h\to0^+}\dfrac1{2+h}=?$$
A: I nice approach using triangles. Use the following figure, where $\triangle ABC$ is right-angled with $\overline{BC} = x>0$ and $\overline{AB} = 1+x$. Take $D$ on $AB$ so that $\overline{AD} = 1$ and $E$ on $AC$ so that $DE\perp AC$. 
 
By definition
$$\angle CAB = \arctan\left(\frac{x}{x+1}\right),$$
and by the External Angle Theorem
$$\angle ACD = \frac{\pi}4-\angle CAB.$$
Note that $\overline{AC} = \sqrt{2x^2+2x+1}$, and use the fact that $\triangle ADE \sim \triangle ABC$ to conclude that
$$\overline{DE} = \frac{x}{\sqrt{2x^2+2x+1}},$$
and
$$\overline{AE} = \frac{x+1}{\sqrt{2x^2+2x+1}},$$
so that
$$\overline{CE} = \frac{2x^2+x}{\sqrt{2x^2+2x+1}},$$
and
$$\angle ACD = \arctan\frac{\overline{DE}}{\overline{CE}}=\arctan\left(\frac{1}{2x+1}\right).$$
So, as per the other answers, your limit is equal to
$$\lim_{x\to +\infty} x\arctan\left(\frac1{2x+1}\right) = \frac12.$$
