Limit without L’Hopital $\lim_{x\to0}\frac{\pi – 4\arctan{1\over 1+x}}{x}$

Question

Evaluate the limit:
$$
\lim_{x\to0}\frac{\pi – 4\arctan{1\over 1+x}}{x}
$$

I've been able to show the limit is equal to $2$ using L'Hopital's rule. After finding the derivative of the nominator the limits simply becomes:
$$
\lim_{x\to0}\frac{4}{x^2 + 2x+2} = 2
$$

I'm looking for a way to find the limit without involving derivatives, but rather using some elementary methods. I've also played around with the identities involving $\arctan x$ but didn't find anything suitable.

Could someone please suggest a method to solve that problem?

Answer 1

Set $\dfrac\pi4-\arctan\dfrac1{1+x}=y$

$\dfrac1{1+x}=\tan\left(\dfrac\pi4-y\right)=\dfrac{1-\tan y}{1+\tan y}$

$x=?$