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June 8th, 2019, 08:33 AM   #1
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How to solve the limit

$\displaystyle \lim_{n\rightarrow \infty } \frac{\sqrt[n]{n}}{\sqrt[2n]{2n}}$.
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June 8th, 2019, 08:59 AM   #2
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$\displaystyle \lim\frac{n^{\frac{1}{n}}}{2^{\frac{1}{2n}} n^{\frac{1}{2n}}} = \lim\frac{n^{\frac{1}{n} - \frac{1}{2n}}}{2^{\frac{1}{2n}}}$

$\lim 2^{\frac{1}{2n}} = 1$ so we have

$\displaystyle \lim n^{\frac{1}{2n}} = e^{ \lim \frac{ln(n)}{2n}}$

Use lhopitals (probably mispelled)

$\displaystyle \lim \frac{ln(n)}{2n} = \lim \frac{ 1}{2n} = 0$

So, we have $ e^{ \lim \frac{ln(n)}{2n}} = 1 $ and therefore the whole limit is $1$
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Last edited by Greens; June 8th, 2019 at 09:01 AM.
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June 8th, 2019, 09:03 AM   #3
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establish that

$\lim \limits_{n\to \infty}\sqrt[n]{\dfrac n c}=1,~\forall c>0$

and the rest is trivial
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June 9th, 2019, 08:30 AM   #4
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Originally Posted by Greens View Post
$\lim 2^{\frac{1}{2n}} = 1$ so we have

$\displaystyle \lim n^{\frac{1}{2n}} = e^{ \lim \frac{ln(n)}{2n}}$
This is invalid isn't it? You can only decompose the limit into two parts like that if you have proved that the limits of both parts exist.
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June 9th, 2019, 10:18 AM   #5
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$\lim \limits_{n\to \infty} \dfrac{\sqrt[n]{n}}{\sqrt[2n]{2n}} = \\

\lim \limits_{n\to \infty} \dfrac{\sqrt[2n]{n^2}}{\sqrt[2n]{2n}} = \\

\lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n^2}{2n}} = \\

\lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n}{2}} = \\

\lim \limits_{n\to \infty} \sqrt[n]{\dfrac{n}{4}}$

consider $\sqrt[n]{\dfrac{n}{c}},~c>0$

$\ln \left(\sqrt[n]{\dfrac n c}\right) = \dfrac 1 n\left(\ln(n) - \ln(c)\right)$

$\lim \limits_{n\to \infty} \ln \left(\sqrt[n]{\dfrac n c}\right) =
\lim \limits_{n\to \infty} \dfrac 1 n\left(\ln(n) - \ln(c)\right) = 0$

(I waved my hands that the first term goes to zero but it's clearly so)

By the substitution theorem for limits

$\lim \limits_{x\to \infty} \sqrt[n]{\dfrac n c} = \\

\exp\left(\lim \limits_{x\to \infty} \ln \left(\sqrt[n]{\dfrac n c}\right)\right) = e^0 = 1$

and so

$\lim \limits_{n\to \infty} \dfrac{\sqrt[n]{n}}{\sqrt[2n]{2n}} =

\lim \limits_{n\to \infty} \sqrt[n]{\dfrac{n}{4}} = 1$
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