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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.

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$a = (\sqrt[n]{a})^n$. If the root is below 1 then $a$ is even smaller, thus any root of a number above 1 is also above 1. It should be intuitively clear that roots approach 1. – Karolis Juodelė Nov 18 '12 at 16:19
$\sqrt[n]{n^2} \gt 1$ for $n\gt 1$. Similarly $\frac{1}{n}\log(n^2) \gt 0$ for $n\gt 1$ since $\frac{1}{n} \gt 0$ and $\log(n^2) = 2\log(n) \gt 0$. – Henry Nov 18 '12 at 16:27
up vote 1 down vote accepted

$$\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\frac{4x^2}{\lim_{n\to \infty}n^{\frac2n}}$$

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, $$\lim_{n \rightarrow \infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\frac{4x^2}{\lim_{n\to \infty}n^{\frac2n}}=4x^2 $$

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And this is related with my limit? =S :feel confused: – Pizzirani Leonardo Nov 18 '12 at 16:09
@PizziraniLeonardo, could you please look into the edited answer. – lab bhattacharjee Nov 18 '12 at 16:13
Why if $lim_{n\rightarrow\infty}ln(m)=0 \implies lim_{n\rightarrow\infty}m=1$ ? – Pizzirani Leonardo Nov 18 '12 at 16:19
Thanks to @user49685 I've understand! $$m=e^{ln(m)}$$ $$ln(m)\rightarrow0 \implies m\rightarrow e^0=1$$ Thanks at all! – Pizzirani Leonardo Nov 18 '12 at 16:22

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 $\sqrt[n]{n^2} \rightarrow 1$.

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$$\lim_{n\to\infty}\sqrt[n]{\frac{4^nx^{2n}}{n^2}}=\lim_{n\to \infty}\left({\frac{4^nx^{2n}}{n^2}}\right)^{\frac{1}{n}}=\lim_{n\to \infty}{\frac{4x^{2}}{n^{\frac{2}{n}}}}=\frac{4x^2}{1}=4x^2$$ because $$\lim_{n\to \infty}n^{\frac{2}{n}}=\lim_{n\to \infty}e^{\log{ n^{2/n}}}=\lim_{n\to \infty}e^{\frac{2log n}{n}}=e^{\lim_{n\to\infty}\frac{2log n}{n}}=e^0=1$$

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Yup, my dubt was about $lim_{n\rightarrow\infty}n^{\frac{2}{n}}=1$ =) – Pizzirani Leonardo Nov 18 '12 at 16:33
that is proved by user 49685 – Adi Dani Nov 18 '12 at 16:36

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