<s>[INST] 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?
 [/INST] 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
$$\lim_{x\to +\infty} x\arctan\left(\frac1{2x+1}\right) = \frac12.$$
</s>