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$$ \lim_{x\to 0} \Big(\frac{\arctan(x)}{x}\Big)^{1/x^2}$$

It should use l'hopital's rules. but I am not sure what to do after putting it to e

Mark Viola
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M.Mass
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1 Answers1

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Note that $\arctan(x)=x-\frac13 x^3+O(x^5)$. Hence, we see that

$$\frac{\arctan(x)}{x}=1-\frac13x^2+O(x^4)$$

Then, letting $1/t=x^2$ we have

$$\begin{align} \lim_{x\to 0}\left(\frac{\arctan(x)}{x}\right)^{1/x^2}&=\lim_{t\to \infty}\left(1-\frac{1/3}{t}+O(1/t^2)\right)^{t}\\\\ &=\lim_{t\to \infty}\left(1-\frac{1/3}{t}\right)^{t}\underbrace{\left(1+O(1/t^2)\right)^{t}}_{\to 1}\\\\ &=e^{-1/3} \end{align}$$

Mark Viola
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