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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) $$ \lim \limits_{x\to 0}\left( \frac{e^{4}\left( 1-x^{2}\right) ^{\frac{1}{x^{2}} }}{\left( 1-2x\right) ^{\frac{1}{x} }} \right) ^{\frac{1}{x} } $$ 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) $$\lim_{n\rightarrow \infty}(\frac{\arcsin(\frac{1}{n})+\arcsin(\frac{2}{n})+...+\arcsin(\frac{n}{n})}{n})$$

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. Thank you in advance.

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    $\begingroup$ The second is a Riemann sum. $\endgroup$ – David Mitra Aug 4 '16 at 10:00
  • $\begingroup$ In the first limit use $\lim_{t\to 0}(1+at)^{1/t}=e^a$. By the way, in the first one you have $x\to 0^+$? $\endgroup$ – Robert Z Aug 4 '16 at 10:08
  • $\begingroup$ For the first one, are you allowed to use Taylor series ? $\endgroup$ – Claude Leibovici Aug 4 '16 at 10:14
  • $\begingroup$ on the first one I was looking at taking log, then putting x^3 in the denominator with x^3 multiplied through the top as needed, then you'd need to differentiate 3 times I think $\endgroup$ – Cato Aug 4 '16 at 10:20
  • $\begingroup$ @RobertZ it's just $x\rightarrow 0$ $\endgroup$ – MathIsTheWayOfLife Aug 4 '16 at 10:33
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For the first if $x \to 0$ then the limit does not exist. To prove this let us observe that $$(1 - x^{2})^{1/x^{2}} \to e^{-1}, (1 - 2x)^{1/x} \to e^{-2}$$ as $x \to 0$ and hence $$f(x) = \left(\frac{e^{4}(1 - x^{2})^{1/x^{2}}}{(1 - 2x)^{1/x}}\right) \to e^{5} > 1$$ and $g(x) = 1/x \to \infty$ as $x \to 0^{+}$ so $\{f(x)\}^{g(x)} \to \infty$ as $x \to 0^{+}$.

But when $x \to 0^{-}$ then $g(x) \to -\infty$ and hence $\{f(x)\}^{g(x)} \to 0$ as $x \to 0^{-}$. It follows that the desired limit does not exist. In the above we have used the following standard limits $$\lim_{x \to 0}(1 + ax)^{1/x} = e^{a},a^{x} \to \infty\text{ as }x \to \infty\text{ if }a > 1$$

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

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Second one: $$L=\lim_{n\rightarrow \infty}(\frac{arcsin(\frac{1}{n})+arcsin(\frac{2}{n})+...+arcsin(\frac{n}{n})}{n}),$$ $$L=\int_0^1 \arcsin{x}dx=(x\arcsin{x}+\sqrt{1-x^2})|_0^1=\frac{\pi-4}{4}.$$

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As regards the first limit, by using $\lim_{t\to 0}(1+at)^{1/t}=\lim_{s\to +\infty}(1+a/s)^{s}=e^a$ we have that $$\lim \limits_{x\to 0}\left( \frac{e^{4}\left( 1-x^{2}\right) ^{\frac{1}{x^{2}} }}{\left( 1-2x\right) ^{\frac{1}{x} }} \right)=\frac{e^4\cdot e^{-1}}{e^{-2}}=e^5.$$ So the base of the exponential is eventually greater then $2$ (something $>1$ and less than $e^5$).

If $x\to 0^+$ then for $0<x<r$ for some $r>0$, $$+\infty \leftarrow 2^{1/x}<\left( \frac{e^{4}\left( 1-x^{2}\right) ^{\frac{1}{x^{2}} }}{\left( 1-2x\right) ^{\frac{1}{x} }} \right) ^{\frac{1}{x} }$$

If $x\to 0^-$ then for $-r<x<0$ 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!

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Just for your curiosity.

Since you have received nice and simple answers for the first problem, let me show what could be done with Taylor series.

Considering $$A=\left( \frac{e^{4}\left( 1-x^{2}\right) ^{\frac{1}{x^{2}} }}{\left( 1-2x\right) ^{\frac{1}{x} }} \right) ^{\frac{1}{x} }$$ $$\log(A)=\frac 1x\,\log\left( \frac{e^{4}\left( 1-x^{2}\right) ^{\frac{1}{x^{2}} }}{\left( 1-2x\right) ^{\frac{1}{x} }} \right)=\frac 1x\,\left(4+\frac 1{x^2}\log(1-x^2)- \frac 1x \log(1-2x)\right)$$ Now, using the fact that around $y=0$ $$\log(1-y)=-y-\frac{y^2}{2}+O\left(y^3\right)$$ we then have $$\frac 1{x^2}\log(1-x^2)=-1-\frac{x^2}{2}+O\left(x^3\right)$$ $$\frac 1x \log(1-2x)=-1-x-\frac{4 x^2}{3}+O\left(x^3\right)$$ All the above makes $$\log(A)=\frac 1x\,\left(4+x+\frac{5 x^2}{6}+O\left(x^3\right)\right)=\frac{4}{x}+1+\frac{5 x}{6}+O\left(x^2\right)$$ So, when $x\to 0^+$, $\log(A)\to \infty$ and $A\to \infty$

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Second limit:

$$\lim_{n\to\infty}(\frac{arcsin(\frac{1}{n})+arcsin(\frac{2}{n})+...+arcsin(\frac{n}{n})}{n}) = 0$$

Although being quite tricky and taking some time for me to wrap my head around, you can actually use L'Hospitals rule on the second limit because its already in $$\lim_{x\to\infty}=\frac{\infty}{\infty}$$ since the limit of the bottom is seen to be infinity and the limit of the top $$\lim_{x\to\infty}\sin(1/x)+\sin(2/x)....$$ also approaches infinity because this is just adding a bunch of numbers between 0 and 1 and the greater value of x the more numbers it adds between 1 and 0 and approaches infinity. From this we can create the series for the top of the fraction in the limit: $$\lim_{x\to\infty}\sum_{n=1}^x\arcsin(\frac{n}{x})$$ From here we apply L'Hopitals. To take the derivative of the sum you just take the derivative of the formula which is the same as taking the derivative of each value in the sum. Our limit after L'Hospital becomes: $$\lim_{x\to\infty}\frac{\sum_{n=1}^x -\frac{n}{x^2\sqrt{(1-\frac{n^2}{x^2})}}}{1}$$ from this take the limit of the first value $$\lim_{x\to\infty}-\frac{1}{x^2\sqrt{1-\frac{n^2}{x^2}}} = 0 $$and last value which n = x so $$\lim_{x\to\infty}-\frac{x}{x^2\sqrt{1-\frac{n^2}{x^2}}} = -\frac{1}{x\sqrt{1-\frac{n^2}{x^2}}} = \lim_{x\to\infty}\frac{1}{x} = 0$$ The limit of the start and end of the sum is 0 and every limit value in between that has to be in between the start and end value since n is increasing linearly so every value is 0. Therefore the limit finally is simplified to top and bottom limit as $$\frac{0}{1}$$ so the limit can be concluded to equal zero.

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Hint: For #1, substitute $t=1/x$ then use the limit definition of $e^x$. For #2, it is a riemann sum.

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  • $\begingroup$ what do you mean with limit definition of $e^x$ ? $\endgroup$ – MathIsTheWayOfLife Aug 4 '16 at 10:19
  • $\begingroup$ $$\lim_{t\to\infty}(1+k/t)^t=e^t$$ $\endgroup$ – Cyclohexanol. Aug 4 '16 at 11:27
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    $\begingroup$ @MathIsTheWayOfLife see above $\endgroup$ – Cyclohexanol. Aug 4 '16 at 11:28

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