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 June 8th, 2019, 08:33 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 How to solve the limit $\displaystyle \lim_{n\rightarrow \infty } \frac{\sqrt[n]{n}}{\sqrt[2n]{2n}}$.
 June 8th, 2019, 08:59 AM #2 Member     Joined: Oct 2018 From: USA Posts: 89 Thanks: 61 Math Focus: Algebraic Geometry $\displaystyle \lim\frac{n^{\frac{1}{n}}}{2^{\frac{1}{2n}} n^{\frac{1}{2n}}} = \lim\frac{n^{\frac{1}{n} - \frac{1}{2n}}}{2^{\frac{1}{2n}}}$ $\lim 2^{\frac{1}{2n}} = 1$ 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 $1$ Thanks from topsquark and idontknow Last edited by Greens; June 8th, 2019 at 09:01 AM.
 June 8th, 2019, 09:03 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 establish that $\lim \limits_{n\to \infty}\sqrt[n]{\dfrac n c}=1,~\forall c>0$ and the rest is trivial Thanks from idontknow
June 9th, 2019, 08:30 AM   #4
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Quote:
 Originally Posted by Greens $\lim 2^{\frac{1}{2n}} = 1$ 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.

 June 9th, 2019, 10:18 AM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 $\lim \limits_{n\to \infty} \dfrac{\sqrt[n]{n}}{\sqrt[2n]{2n}} = \\ \lim \limits_{n\to \infty} \dfrac{\sqrt[2n]{n^2}}{\sqrt[2n]{2n}} = \\ \lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n^2}{2n}} = \\ \lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n}{2}} = \\ \lim \limits_{n\to \infty} \sqrt[n]{\dfrac{n}{4}}$ 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 $\lim \limits_{n\to \infty} \dfrac{\sqrt[n]{n}}{\sqrt[2n]{2n}} = \lim \limits_{n\to \infty} \sqrt[n]{\dfrac{n}{4}} = 1$ Thanks from topsquark and idontknow

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