Using $\ln x<(x^\epsilon-1)/\epsilon$ for $x>1$ and $\epsilon>0$, and the squeeze theorem, show that $\lim\limits_{x\to\infty}(\ln x)/x^\epsilon=0$

Question

A question from Introduction to analysis by Arthur Mattuck:

Given $\epsilon \gt 0$, $\ln x \lt \frac {(x^\epsilon - 1)} {\epsilon}$, for $x \gt 1$. Deduce that $\lim_{x \to \infty} \frac{\ln x}{x^\epsilon}=0$ for any $\epsilon \gt 0$ using the squeeze theorem.

Now I know how to prove the first inequality by integration, but how to get the second limit by using Squeeze Theorem? $$\frac{\ln x}{x^\epsilon} \lt \frac{(x^\epsilon - 1)}{\epsilon x^\epsilon}=\frac{1}{\epsilon}(1-\frac{1}{x^\epsilon})$$ The limit of $\frac{1}{\epsilon}(1-\frac{1}{x^\epsilon})$ is $\frac{1}{\epsilon}$ as $x$ tends to $\infty$, not $0$.

Answer 1

hint. $$\ln x=\int_1^x\frac{1}{t}\,dt<\int_1^x\frac{t^\varepsilon}{t}\,dt=\frac{x^\varepsilon-1}{\varepsilon}$$