June 8th, 2019, 08:33 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81  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: 63 Thanks: 45 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 $1$ 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: 2,452 Thanks: 1337 
establish that $\lim \limits_{n\to \infty}\sqrt[n]{\dfrac n c}=1,~\forall c>0$ and the rest is trivial 
June 9th, 2019, 08:30 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra  
June 9th, 2019, 10:18 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
$\lim \limits_{n\to \infty} \dfrac{\sqrt[n]{n}}{\sqrt[2n]{2n}} = \\ \lim \limits_{n\to \infty} \dfrac{\sqrt[2n]{n^2}}{\sqrt[2n]{2n}} = \\ \lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n^2}{2n}} = \\ \lim \limits_{n\to \infty} \sqrt[2n]{\dfrac{n}{2}} = \\ \lim \limits_{n\to \infty} \sqrt[n]{\dfrac{n}{4}}$ 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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