# So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one.

So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one. 1)
$\underset{x\to 0}{lim}{\left(\frac{{e}^{4}{\left(1-{x}^{2}\right)}^{\frac{1}{{x}^{2}}}}{{\left(1-2x\right)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$
I am having problems with this one to make it into one that can be used for L'Hospital's rule, but i cannot get it into $\frac{0}{0}$, there are others options as well, but i am stuck.
2) $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{arcsin}\left(\frac{1}{n}\right)+\mathrm{arcsin}\left(\frac{2}{n}\right)+...+\mathrm{arcsin}\left(\frac{n}{n}\right)}{n}\right)$
So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L'Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.
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Isaiah Haynes
Step 1
For the first if $x\to 0$ then the limit does not exist. To prove this let us observe that
$\left(1-{x}^{2}{\right)}^{1/{x}^{2}}\to {e}^{-1},\left(1-2x{\right)}^{1/x}\to {e}^{-2}$
as $x\to 0$ and hence
$f\left(x\right)=\left(\frac{{e}^{4}\left(1-{x}^{2}{\right)}^{1/{x}^{2}}}{\left(1-2x{\right)}^{1/x}}\right)\to {e}^{5}>1$
and $g\left(x\right)=1/x\to \mathrm{\infty }$ as $x\to {0}^{+}$ so $\left\{f\left(x\right){\right\}}^{g\left(x\right)}\to \mathrm{\infty }$ as $x\to {0}^{+}$.
Step 2
But when $x\to {0}^{-}$ then $g\left(x\right)\to -\mathrm{\infty }$ and hence $\left\{f\left(x\right){\right\}}^{g\left(x\right)}\to 0$ as $x\to {0}^{-}$. It follows that the desired limit does not exist. In the above we have used the following standard limits

The second limit is a Riemann sum and can be evaluated easily (as explained in answer from alans).

yamyekay3
Step 1
As regards the first limit, by using
$\underset{t\to 0}{lim}\left(1+at{\right)}^{1/t}=\underset{s\to +\mathrm{\infty }}{lim}\left(1+a/s{\right)}^{s}={e}^{a}$ we have that
$\underset{x\to 0}{lim}\left(\frac{{e}^{4}{\left(1-{x}^{2}\right)}^{\frac{1}{{x}^{2}}}}{{\left(1-2x\right)}^{\frac{1}{x}}}\right)=\frac{{e}^{4}\cdot {e}^{-1}}{{e}^{-2}}={e}^{5}.$
So the base of the exponential is eventually greater then 2 (something $>1$ and less than ${e}^{5}$).
Step 2
If $x\to {0}^{+}$ then for $0 for some $r>0$,
$+\mathrm{\infty }←{2}^{1/x}<{\left(\frac{{e}^{4}{\left(1-{x}^{2}\right)}^{\frac{1}{{x}^{2}}}}{{\left(1-2x\right)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$
If $x\to {0}^{-}$ then for $-r for some $r>0$,
$0<{\left(\frac{{e}^{4}{\left(1-{x}^{2}\right)}^{\frac{1}{{x}^{2}}}}{{\left(1-2x\right)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}<{2}^{1/x}\to 0.$
Therefore you have different behaviours for $x\to {0}^{-}$ and $x\to {0}^{+}$ and the limit for $x\to 0$ does not exist!