I'm trying to find out why: $\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}} = 4x^2$ Seems to me that it go $\rightarrow\infty$ because of the $\sqrt[n]{n^2}\rightarrow_{n\rightarrow\infty}0$. What I'm doing wrong?
Thanks! Leonardo.
I'm trying to find out why: $\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}} = 4x^2$ Seems to me that it go $\rightarrow\infty$ because of the $\sqrt[n]{n^2}\rightarrow_{n\rightarrow\infty}0$. What I'm doing wrong?
Thanks! Leonardo.
$\lim_{n\to\infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\lim_{n\to \infty}\left({\frac{4^nx^{2n}}{n^2}}\right)^{\frac{1}{n}}=\lim_{n\to \infty}{\frac{4x^{2}}{n^{\frac{2}{n}}}}=\frac{4x^2}{1}=4x^2$ because $\lim_{n\to \infty}n^{\frac{2}{n}}=\lim_{n\to \infty}e^{\log{ n^{2/n}}}=\lim_{n\to \infty}e^{\frac{2log n}{n}}=e^{\lim_{n\to\infty}\frac{2log n}{n}}=e^0=1$
No, that's NOT true.
$\lim \sqrt[n]{n} = 1$, you can check this by plugging some big $n$ to the calculator, say $n = 10000$, then try to calculate $\sqrt[10000]{10000}$, it'll be close to 1.
Let $y = \sqrt[n]{n}$, since $y = e^{\ln y}$, so $\lim \limits_{n \rightarrow \infty} y = \lim \limits_{n \rightarrow \infty} \lim e^{\ln y} = \lim e^{\lim \limits_{n \rightarrow \infty} \ln y}$. We'll now calculate $\lim \limits_{n \rightarrow \infty} \ln y$, then raise $e$ to our result, and get the desired answer.
$\lim \limits_{n \rightarrow \infty} \ln y = \lim \limits_{n \rightarrow \infty} \ln \sqrt[n]{n} = \lim \limits_{n \rightarrow \infty} \ln n^{\frac{1}{n}} = \lim \limits_{n \rightarrow \infty} \frac{1}{n} \ln n = \lim \limits_{n \rightarrow \infty} \frac{\ln n}{n} \mathop{=}\limits^{\mbox{L'Hopital}} \lim \limits_{n \rightarrow \infty} \frac{\frac{1}{n}}{1} = 0$.
So $y \rightarrow e^0 = 1$, or in other words, $\sqrt[n]{n} \rightarrow 1$, which then implies $\sqrt[n]{n^2} \rightarrow 1$.
$\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\frac{4x^2}{\lim_{n\to \infty}n^{\frac2n}}$
Let $m=n^{\frac2n}$
$\log m=2\frac{\log n}n$
As $\lim_{n\to \infty}\frac{\log n}n$ is of the form $\frac{\infty}{\infty},$
we can apply L'Hospital Rule, $\lim_{n\to \infty}\frac{\log n}n=\lim_{n\to \infty}\frac1n=0$
So, $\lim_{n\to \infty}\log m=0\implies \lim_{n\to \infty}m=1$ $\implies \lim_{n\to \infty}n^{\frac2n}=1$
So, $\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\frac{4x^2}{\lim_{n\to \infty}n^{\frac2n}}=4x^2 $