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

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

$x=?$

Answer 2

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 $-4\,f'(0)=2$

Answer 3

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*}