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

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?

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

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

$$x=?$$

• This one is the most elegant so far. Thank you! – roman Apr 11 '19 at 16:30

You may use two facts:

• $$\arctan a - \arctan b = \arctan \frac{a-b}{1+ab}$$
• $$\lim_{y\to 0}\frac{\arctan y}{y} = \lim_{t\to 0}\frac{t}{\tan t} = 1$$

$$\begin{eqnarray*}\frac{\pi - 4\arctan{1\over 1+x}}{x} & = & 4\cdot \frac{\frac{\pi}{4} - \arctan\frac{1}{1+x}}{x}\\ & = & 4\cdot \frac{\arctan \frac{1-\frac{1}{1+x}}{1+\frac{1}{1+x}}}{x} \\ & = & 4\cdot \frac{\arctan \frac{x}{x+2}}{\frac{x}{x+2}\cdot (x+2)} \\ & \stackrel{x \to 0}{\longrightarrow} &\frac{4}{2} = 2 \end{eqnarray*}$$

• Nice approach. Thank you! Definitely (+1) – roman Apr 11 '19 at 16:33
• You are welcome and thank you. :-) – trancelocation Apr 11 '19 at 16:33

Rewrite as $$-4\frac{\arctan(1/(1+x))-\pi/4}{x}$$ then it is of the form: $$-4\,(f(x)-f(0))/(x-0)$$ with $$f(x)=\arctan(1/(1+x))$$ which by definition tends to $$-4\,f'(0)=2$$

• To my mind, your answer actually shows that using Hospital's rule is circular, as the original limit to evaluate is a derivative. On the other hand, the original question explicitly asked for no derivatives at all... – peter a g Apr 11 '19 at 16:13