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How do I solve:

$$
lim_{x to 0}left[1 - xarctanleft(nxright)right]^{, 1/x^{2}}
$$

I know the answer is $e^{-n}$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?

Thanks.

You can do it the following way:

$$
lim_{xto 0 } (1-xarctan(nx))^{1/x^2} = lim_{xto 0 } e^{log((1-xarctan(nx))^{1/x^2})}
$$

doing some algebra on the exponent:
$$
lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
$$

by limit rules:

$$
lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
$$

apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
$$

simplifying

$$
expleft(-frac{n+n+n^2}{2}right) = e^{-n}
$$

Hint: $lim_{xto 0} (1-nx)^{(1/x)}=e^{-n}$

Also look at the Taylor expansion of the arctan

Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have

$$(1-xarctan nx)^{1/x^2}=((1-xarctan nx)^{1/(xarctan nx)})^{(arctan nx)/x}=((1-u)^{1/u})^{(arctan nx)/x}to (e^{-1})^n=e^{-n}$$

using the general limit property $lim f(x)^{g(x)}=(lim f(x))^{lim g(x)}$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.