# Why is this limit: $\lim _{x\to -1+}\frac{\left(-\frac{2}{\pi }x^2\arctan \left(\frac{1}{x+1}\right)+1\right)}{x+1}$ equals to $2+\frac{2}{\pi}$?

I tried to solve this limit since yesterday, and always I found the result is $2$, but when I put it in wolf site, the result was $2+2/p$ why is that?

The limit is : $\lim _{x\to -1+} \frac{\left(-\frac{2}{\pi }x^2\arctan \left(\frac{1}{x+1}\right)+1\right)}{x+1}$ I tried to replace the $\arctan(1/x+1)$ with $\pi/2 - \arctan(x+1)$ since $t\gt0$: $\arctan(t)+\arctan(1/t)=\pi/2$

Can you help me to figure out what was my mistake? Please do no use the L'Hopital Rule, it's not allowed ! And Thanks in advance.

• what is $p$, is it $\pi$? :) Dec 4, 2015 at 15:11
• @H.R. Yes it is. :) Dec 6, 2015 at 9:18

Put $h = x + 1$ and now we have $h \to 0+$. We then obtain $$\frac{\left(-\frac{2}{\pi }x^2\arctan \left(\frac{1}{x+1}\right)+1\right)}{x+1} = \frac {-\frac {2} {\pi} (h - 1)^2 \arctan (1/h) + 1} {h} \to \frac {-\frac {2} {\pi} (h - 1)^2 \frac {\pi} {2} + 1} {h} = \frac {1 - (h - 1)^2} {h} = 2 - h \to 2.$$ So your initial result was correct.

I Found the answer ! I should change the variable in the end so the limit will be like this:

$$\lim _{t\to 0+}-t+1+1+\frac{\left(\frac{2}{\pi }\left(t-1\right)^2\arctan \left(t\right)\right)}{t}$$

And the result is $2 +2/ \pi$ :)

Note: $t=x+1$, that means $x=t-1$

• You should edit the question in spite of writing such an answer.
– user249332
Dec 4, 2015 at 15:03
• @SubhadeepDey Why? can you suggest the edit please? Dec 4, 2015 at 15:04
• I was trying to say that, include this answer in the OP. That will fetch more attention and upvotes towards your OP.
– user249332
Dec 4, 2015 at 15:13
• (+1) for your effort. $0_+$ OR $0_-$, I think you should mention that in your answer. :) Dec 6, 2015 at 9:35
• @H.R. You're right, thanks I'll edit it now. And Thanks again for your +1 :) Dec 7, 2015 at 20:13