Julius Blankenship

2022-09-14

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)

$$\underset{x\to 0}{lim}{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$

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) $\underset{n\to \mathrm{\infty}}{lim}(\frac{\mathrm{arcsin}(\frac{1}{n})+\mathrm{arcsin}(\frac{2}{n})+...+\mathrm{arcsin}(\frac{n}{n})}{n})$

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.

$$\underset{x\to 0}{lim}{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$

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) $\underset{n\to \mathrm{\infty}}{lim}(\frac{\mathrm{arcsin}(\frac{1}{n})+\mathrm{arcsin}(\frac{2}{n})+...+\mathrm{arcsin}(\frac{n}{n})}{n})$

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.

Isaiah Haynes

Beginner2022-09-15Added 16 answers

Step 1

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 $g(x)=1/x\to \mathrm{\infty}$ as $x\to {0}^{+}$ so $\{f(x){\}}^{g(x)}\to \mathrm{\infty}$ as $x\to {0}^{+}$.

Step 2

But when $x\to {0}^{-}$ then $g(x)\to -\mathrm{\infty}$ 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

$$\underset{x\to 0}{lim}(1+ax{)}^{1/x}={e}^{a},{a}^{x}\to \mathrm{\infty}\text{as}x\to \mathrm{\infty}\text{if}a1$$

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

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 $g(x)=1/x\to \mathrm{\infty}$ as $x\to {0}^{+}$ so $\{f(x){\}}^{g(x)}\to \mathrm{\infty}$ as $x\to {0}^{+}$.

Step 2

But when $x\to {0}^{-}$ then $g(x)\to -\mathrm{\infty}$ 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

$$\underset{x\to 0}{lim}(1+ax{)}^{1/x}={e}^{a},{a}^{x}\to \mathrm{\infty}\text{as}x\to \mathrm{\infty}\text{if}a1$$

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

yamyekay3

Beginner2022-09-16Added 1 answers

Step 1

As regards the first limit, by using

$\underset{t\to 0}{lim}(1+at{)}^{1/t}=\underset{s\to +\mathrm{\infty}}{lim}(1+a/s{)}^{s}={e}^{a}$ we have that

$$\underset{x\to 0}{lim}\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)=\frac{{e}^{4}\cdot {e}^{-1}}{{e}^{-2}}={e}^{5}.$$

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 $0<x<r$ for some $r>0$,

$$+\mathrm{\infty}\leftarrow {2}^{1/x}<{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$

If $x\to {0}^{-}$ then for $-r<x<0$ for some $r>0$,

$$0<{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}<{2}^{1/x}\to 0.$$

Therefore you have different behaviours for $x\to {0}^{-}$ and $x\to {0}^{+}$ and the limit for $x\to 0$ does not exist!

As regards the first limit, by using

$\underset{t\to 0}{lim}(1+at{)}^{1/t}=\underset{s\to +\mathrm{\infty}}{lim}(1+a/s{)}^{s}={e}^{a}$ we have that

$$\underset{x\to 0}{lim}\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)=\frac{{e}^{4}\cdot {e}^{-1}}{{e}^{-2}}={e}^{5}.$$

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 $0<x<r$ for some $r>0$,

$$+\mathrm{\infty}\leftarrow {2}^{1/x}<{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$

If $x\to {0}^{-}$ then for $-r<x<0$ for some $r>0$,

$$0<{\left(\frac{{e}^{4}{(1-{x}^{2})}^{\frac{1}{{x}^{2}}}}{{(1-2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}<{2}^{1/x}\to 0.$$

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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