How to prove $\lim\limits_{n\to\infty}\frac{n^{1000}}{n^{\sqrt{n}}}=0$?

Question

$\lim\limits_{n\to\infty}\frac{n^{1000}}{n^{\sqrt{n}}}=0$

I mean it's kind of obvious, but how do I prove it correctly?

Answer 1

Here is a small hint: if $n > 1001^2$ then $\displaystyle \frac{n^{1000}}{n^{\sqrt{n}}} < \frac{1}{n}$.

Answer 2

Hint. If $n\geq 1001^2$ then $$0\leq \frac{n^{1000}}{n^{\sqrt{n}}}\leq \frac{n^{1000}}{n^{1001}}=\frac{1}{n}.$$

Answer 3

Try writing it in exponent form $$ \frac{n^{1000}}{n^{\sqrt{n}}} = \frac{e^{1000 \ln(n)}}{e^{\sqrt{n} \ln(n)}} = e^{1000 \ln(n) - \sqrt{n} \ln(n)} = e^{(1000 - \sqrt{n}) \ln(n)} $$

Answer 4

Prove that if $0 < \epsilon$, then for all $n > \max(2000000, 1/\epsilon)$ that $|f(n)| < \epsilon$.