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The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's.

Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up.

Here is the evaluation using l'hospital's in case someone visits with this question: $$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\ \Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$ and by l'hospital's $$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\ \Rightarrow L=e$$

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Let $f(x) = 1+\arctan (x/2).$ Apply $\ln $ to the expression of interest to get

$$\frac{\ln (f(x))}{x/2} = 2\frac{\ln f(x) - \ln f(0)}{x-0}.$$

As $x\to 0,$ the last expression $\to 2(\ln f)'(0)$ by definition of the derivative (no L'Hopital used). That's easy enough to compute. Exponentiate back for the original limit.

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  • $\begingroup$ This is so clever, this is why I love asking these questions on stackexchange. Thank you. $\endgroup$ – qbert Jun 16 '16 at 21:48
  • $\begingroup$ You're welcome. Loads of problems students want to sic the L'Hopital dog on are actually derivatives in disguise. Good to remember when you encounter 0/0. $\endgroup$ – zhw. Jun 16 '16 at 22:57
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Rewrite as $$ \lim_{x\rightarrow 0}\exp\left[\frac{2}{x}\ln\left(1+\arctan(\frac{x}{2})\right)\right] $$ and then use $\arctan(x/2)\sim x/2$ for $x\to 0$, and $\ln (1+x/2)\sim x/2$ for $x\to 0$. Therefore the limit is $=\exp[(2/x)(x/2)]=\mathrm{e}$.

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  • $\begingroup$ Ah shoot I should have thought about that. Thank you $\endgroup$ – qbert Jun 16 '16 at 21:03
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From Taylor series, you can get more than just the limit. Consider $$y=\left(1+\arctan\left(\frac{x}{2}\right)\right)^{\frac{2}{x}}$$ and, as usual for this kind of problem, take logarithms $$\log(y)=\frac{2}{x}\,\log\left(1+\arctan(\frac{x}{2}) \right)$$ Now, use the classical $$\arctan(y)=y-\frac{y^3}{3}+O\left(y^4\right)$$ Replace $y$ by $\frac{x}{2}$ which makes $$\arctan(\frac{x}{2})=\frac{x}{2}-\frac{x^3}{24}+O\left(x^4\right)$$ $$\log(y)=\frac{2}{x}\,\log\left(1+\frac{x}{2}-\frac{x^3}{24}+O\left(x^4\right)\right)$$ Using again the expansion of $\log(1+y)$, we end with $$\log(y)=\frac{2}{x}\times\left(\frac{x}{2}-\frac{x^2}{8}+O\left(x^4\right)\right)=1-\frac{x}{4}+O\left(x^3\right)$$ Now, using $y=e^{\log(y)}$ and Taylor again $$y=e\left(1-\frac{x}{4}+\frac{x^2}{32}\right)+O\left(x^3\right)$$

If, for fun, you plot the original function and the last approximation for $0\leq x\leq 1$, you should notice that they almost coincide. So, suppose that you need to solve for $x$ equation $$\left(1+\arctan\left(\frac{x}{2}\right)\right)^{\frac{2}{x}}=2$$ solving the quadratic will give an estimate $$4-4 \sqrt{\frac{4}{e}-1}\approx 1.25331$$ while the exact solution is $\approx 1.33627$.

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First and foremost you can get rid of the $x/2$ (it is a pain to type fractions) by putting $t = x/2$ and using the fact that as $x \to 0$ the variable $t \to 0$. We thus need to evaluate $$L = \lim_{t \to 0}(1 + \arctan t)^{1/t}\tag{1}$$ We can now proceed in your manner \begin{align} \log L &= \log\left\{\lim_{t \to 0}(1 + \arctan t)^{1/t}\right\}\notag\\ &= \lim_{t \to 0}\log(1 + \arctan t)^{1/t}\text{ (via continuity of log)}\notag\\ &= \lim_{t \to 0}\frac{\log(1 + \arctan t)}{t}\notag\\ &= \lim_{t \to 0}\frac{\log(1 + \arctan t)}{\arctan t}\cdot\frac{\arctan t}{t}\notag\\ &= 1\cdot 1\notag\\ &= 1 \notag \end{align} Hence $L = e$.

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$$\lim _{ x\rightarrow 0 } \left( 1+\arctan \left( \frac { x }{ 2 } \right) \right) ^{ \frac { 2 }{ x } }=\lim _{ x\rightarrow 0 }{ \left[ \left( 1+\arctan \left( \frac { x }{ 2 } \right) \right) ^{ \frac { 1 }{ \arctan \left( \frac { x }{ 2 } \right) } } \right] } ^{ \frac { 2 }{ x } \arctan \left( \frac { x }{ 2 } \right) }=\\ =e^{ \lim _{ x\rightarrow 0 } \frac { 2 }{ x } \arctan \left( \frac { x }{ 2 } \right) }=e$$ beacuse it is well known that :$$\lim _{ x\rightarrow 0 } \frac { 2 }{ x } \arctan =\lim _{ x\rightarrow 0 } \frac { \arctan \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } =1$$

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Taylor, first order, would do. $\arctan u = u+o(u)$, $\ln(1+u) = u+o(u)$, so than $\ln(1+\arctan u) = u+o(u)$ when $u\to 0$. So your limit is $e^{\frac{1}{u}\ln(1+\arctan u)} = e^{1+o(1)}$, with $u=\frac{x}{2}$; $$e^{\frac{1}{u}\ln(1+\arctan u)} \xrightarrow[u\to0]{} e^1 = e.$$


Remark: there may be simpler. But this works, and takes roughly one minute if you know the expansions at $0$ (to very low order, here first) of some standard functions.

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