Proving limit without using L'hopital rule

Question

I want to know how to prove $$\underset{n\to\infty}{\lim}\dfrac{(\ln n)^{\alpha}}{n^b}=0\quad\forall b>0$$ without using L'hopital rule or Taylor's expansion. I want to find a proof for high school students.
I know the problem was asked here before Proving $\lim_{n\to \infty} \frac{(\ln n)^a}{n^b} = 0$ for all $a,b>0$, but this does not answer my question.

Answer 1

The answer to the linked question can be adapted using the squeeze theorem.

$$L=\lim\limits_{n\to+\infty}\left(\frac {\log n}{n^{b/\alpha}}\right)^{\alpha}$$

Since $0\leq\log n\leq n$ for $n\geq1$, then dividing both sides by $n^{b/\alpha}$ gives

$$0\leq\frac {\log n}{n^{b/\alpha}}\leq\frac 1{n^{b/\alpha-1}}$$

Taking the limit as $n\to+\infty$, we can see that $L=0$. This only applies for $b>0$.

Answer 2

We just to show $\displaystyle\ln n<n^{c}$ for large $n$ and $c>0$. To get this, let $f(x)=\ln x-x^{c}$, then $$f'(x)=\frac{1}{x}-cx^{c-1}=x^{-1}(1-cx^c),$$ where $1-cx^c<0$ for large $x$. so $f(x)$ decreases for large $x$. Then $f(e^{N})=N-e^{cN}$. It's easy to see $f(e^N)<0$ for large $N$ (Similar to what was done earlier, take the derivative and then figure out the extremes).

Finally, we get $\ln x-x^c<0$ for large $x$. Let $c=\frac{b}{2a}$,then we have $$0<\lim_{n\to \infty}\frac{(\ln n)^a}{n^b}<\lim_{n\to \infty}n^{ac-b}=\lim_{n\to \infty}n^{-\frac{b}{2}}=0.$$