$$ \lim_{x\to 0} \Big(\frac{\arctan(x)}{x}\Big)^{1/x^2}$$
It should use l'hopital's rules. but I am not sure what to do after putting it to e
How to evaluate the limit $ \lim_{x\to 0} \Big(\frac{\arctan(x)}{x}\Big)^{1/x^2}$?
2
$\begingroup$
limits
-
0What is $\operatorname{arctg}x$? Is it $\arctan$? – 2017-02-20
-
0Is it $\arctan x$? Or what? – 2017-02-20
-
0there is a missing argument in $\arctan$ function – 2017-02-20
1 Answers
3
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
$$\begin{align} \lim_{x\to 0}\left(\frac{\arctan(x)}{x}\right)^{1/x^2}&=\lim_{t\to \infty}\left(1-\frac{1/3}{t}+O(1/t^2)\right)^{t}\\\\ &=\lim_{t\to \infty}\left(1-\frac{1/3}{t}\right)^{t}\underbrace{\left(1+O(1/t^2)\right)^{t}}_{\to 1}\\\\ &=e^{-1/3} \end{align}$$