# $\lim\limits_{n\to\infty} \sup \sqrt[n]{n^k}$ for $k \in \mathbb{N}$?

What is $$\lim\limits_{n\to\infty} \sup \sqrt[n]{n^k} \ \ \ \text{where } k \in \mathbb{N}\text ?$$ I've seen the proof that $\lim\limits_{n\to\infty} \sqrt[n]n = 1$. I believe the answer is $1$ for all $k$ because $\sqrt[n]n$ approaches $1$, so a sequence that approaches $1$ raised to an exponent, I believe, should also approach $1$.

I don't know how to prove this, mainly because of the $\lim\limits_{n\to\infty}\sup$ concept. I learned $\lim\limits_{n\to\infty}\sup s_n$ as the $\sup$ of all subsequential limits of $s_n$. Intuitively I know it's the $\inf$ of all $\sup$s of the sequence as $n\to\infty$. But I don't know how to use that definition /intuition to solve problems like this.

• Please use \limsup_{n\to\infty}, not \lim_{n\to\infty}\sup. – Did Jul 8 '15 at 20:22

Observe that $$\sqrt[n]{n^k}=(\sqrt[n]{n})^k.$$ Since $f(x)=x^k$ is continuous, $f(\sqrt[n]{n})\rightarrow f(1)=1$ as $\sqrt[n]{n} \rightarrow 1$ an the result follows.

$$\sqrt[n]{n^k} = \exp( \frac{1}{n} \ln(n^k) ) = \exp( k \frac{\ln(n)}{n} ).$$
Now as $$\frac{ \ln(n) }{n} \to 0$$ and that $e^{kx}$ is continuous, you get that
$$\lim_{n\to + \infty} \exp( k \frac{\ln(n)}{n} ) = e^0 = 1.$$
Hint: if $\lim$ exists, then $\limsup=\lim=\liminf$.