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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?

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

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

$x=?$

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  • $\begingroup$ This one is the most elegant so far. Thank you! $\endgroup$ – roman Apr 11 at 16:30
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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*}

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  • $\begingroup$ Nice approach. Thank you! Definitely (+1) $\endgroup$ – roman Apr 11 at 16:33
  • $\begingroup$ You are welcome and thank you. :-) $\endgroup$ – trancelocation Apr 11 at 16:33
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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 $-4\,f'(0)=2$

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    $\begingroup$ 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... $\endgroup$ – peter a g Apr 11 at 16:13

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