$\frac{\log{x}}{x^{\epsilon}}$ Tends to 0 as x Tends to Infinity?

Question

$\theta(x) :=\sum_{p\leq x}\log{p}$, and $\epsilon>0$. It is written at the end that-

The second term on the RHS tends to 0 as x → ∞, and the lemma follows: by choosing $\epsilon$ sufficiently small we can make the ratios of ϑ(x) to x and π(x) to x/ log x arbitrarily close together as x → ∞, so if one of them tends to 1, then so must the other.

I couldn't understand -

1. How the second term on the RHS tends to 0 as x → ∞?
2. How by choosing $\epsilon$ sufficiently small we can make the ratios of ϑ(x) to x and π(x) to x/ log x arbitrarily close together as x → ∞ ?

The source of the question is below theorem-

Answer 1

By $x=e^y$

$$\frac{\log x}{x^\epsilon}=\frac{y}{e^{\epsilon y}}$$

and for any $\epsilon>0$ eventually $e^{\epsilon y}\ge y^2$.

For the second part since $\frac{\log x}{x^\epsilon}\to 0$ for any $\epsilon >0$, we can make $\frac{\theta(x)}x$ arbitraly close to $\frac{\pi(x)\log x}{x}$.