Using a Macbook Pro as a second monitor for another Macbook Pro

Since $f(x)=x^p$ is continuous at $x=1$, it suffices to show $\lim_{n\to\infty} n^{1/n}=1$. Prove: for any $\delta>0$, there exists $N>0$ such that $n<(1+\delta)^n$ if $n>N$ (Hint: If $n$ is an integer you may use Binomial theorem).

By the l'Hospital theorem we have easily $$\lim_{n\to\infty}\dfrac{\log n}{n}=0$$ hence we find $$\lim_{n\to\infty}\left(n^p\right)^{\frac{1}{n}}=\lim_{n\to\infty}e^{\frac{\log (n^p)}{n}}=\lim_{n\to\infty}e^{p\frac{\log (n)}{n}}=e^0=1$$

Here's a possible way to see this.

One chooses the smallest integer $d$ such that $e-nd<0$. Note that $nd \le e+n$.

Then if $D$ is effective of degree $d$, $h^0(E(-D))=0$ since $E(-D)$ is semistable with negative degree.

Since $h^0(E)\le h^0(E(-D))+nd$ (by using the exact sequence $0\to \mathcal O(-D)\to \mathcal O \to \mathcal O_D\to 0$ tensorized by $E$), the result follows from the inequality $nd\le e+n$ observed above.