Limit with Arctan Here's a hard limit I've been trying to answer for a while : 
$$\lim_{x\rightarrow 1} \dfrac{-2x\arctan{x} + \dfrac{\pi}{2}}{x-1}$$
I've tried all the tricks that the teacher has taught us and still nothing, I even tried to factor the top by $(x-1)$ but still nothing. Can I get some help on how to evaluate this limit please?
 A: Note that if we write $f(x) = -2x\arctan(x)$, then your limit is the formal definition of $f'(1)$. 
Hence you can calculate the limit by calculating that derivative using standard results: $\displaystyle f'(x) = -2\arctan(x)  - \frac{2x}{1+x^2}$, thus ...
A: \begin{align*}
\frac{\frac\pi2 - 2x\arctan x}{x-1}
&= 2 \frac{\frac\pi4 - x\arctan x}{x-1} \\
&= 2\left(\frac{\frac\pi4 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan 1 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan\left(\frac{1-x}{1+x}\right)}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan\left(\frac{1-x}{1+x}\right)}{\left(\frac{1-x}{1+x}\right)}\cdot\frac{-1}{1+x} - \arctan x\right)
\end{align*}
Recall that $\lim_{x\to 0} \frac{\sin x}{x} = 1$; from this it follows that $\lim_{x\to 0} \frac{\tan x}{x} = 1$; from that it follows that $\lim_{x\to 0} \frac{x}{\arctan x} = 1$; and note that $\frac{1-x}{1+x}\to 0$ as $x\to 1$.
A: in cases which have (0/0) or (inf/inf) you can use (Hopital's rule):

so:

and this is easily can be calculated.
