Calculus problem #2603

$$ Let $\,x=\sqrt[n]{n}$ , then $\ln x = \frac{1}{n}\ln n\;\;\;\Rightarrow\;\;\;x=e^{\frac{\ln n}{n}}.$ $$ By using Maclaurin series of exponential function , rewrite $$ $\begin{aligned} \lim_{n\to\infty}\sqrt[n]{n}&= \lim_{n\to\infty} e^{\frac{\ln n}{n}}\\ &=\lim_{n\to\infty}\sum_{k=1}^\infty \frac{\left(\frac{\ln n}{n}\right)^k}{k!}\\ &=\lim_{n\to\infty}\left(1+\frac{\ln n}{n} +O(n^2)\right).\\ \end{aligned}$ $$ It's easy to notice that $\lim\limits_{n\to\infty} \frac{\ln n}{n}=0$ and $\lim\limits_{n\to\infty}O(n^2)=0$ by using L'Hôpital's rule . Thus $$ $\lim_{n\to\infty}\sqrt[n]{n}= \boxed{1}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Since $\sqrt[n]{n} = n^{\frac {1}{n}}$ , you just need to replace $n$ with $\infty$ to get an equation $\infty^{\frac {1}{\infty}}$ . $\frac {1}{\infty}$ equals to $0$ and any number raised to the power to zero is equal to 1 so the answer is 1.

[; \sqrt[n]{n}=n^{1/n} ;] [; n=\infty ;] [; \infty^{1/\infty=0} ;] [; anything^0=1 ;]