Proving $\log n \leq n^\delta$ for $\delta > 0$, $n$ big I am trying to understand this trick:

(You can ignore my handwriting)
WolframAlpha gives that
$\displaystyle\lim_{n \to \infty} \frac {\log n}{n^\delta}=0$. If that is true, can you show me why? If this is the reasoning behind the inequality above, $1$ was probably chosen because the expression is less than $1$ for sufficiently large $n$.
On the other hand, Dahn Jahn has mentioned the inequality $$\log n < n^\delta$$ elsewhere on this site. But I do not know how to prove this. (WolframAlpha doesn't give a clear answer.)
EDIT: I do not need any answers explaining why $\frac{\log n}{n^{\delta}} \to 0$ because Augustin showed me that in the first comment, and I immediately understood. My question is why $1$ was chosen on the RHS of $\frac{\log n}{n^{\delta}}\leq 1$.
 A: Applying l'Hopital, we can easily check that for any $\delta>0$, we have $$\lim_{x\to\infty}\frac{\log x}{x^\delta}\ \overset{l'H}{=}\ \lim_{x\to\infty}\frac{1}{\delta}\cdot\frac{1}{x^\delta}=0.$$ Since this is true for $x\to\infty$, it must be true for $n\to\infty$. In fact, this argument will show any power of the logarithm over any positive power of $x$ will go to $0$.
A: It suffices to prove that for every $\delta > 0$ we have $\log(n)/n^{\delta} \to 0$. Let $a > 0$. If $n > 1$, then
$$
0 < \log n = \int_{t=1}^{n}\frac{1}{t} < \int_{t=1}^{n} t^{a-1} < \frac{n^{a}}{a},
$$
so 
$$
0 < \frac{\log n}{n^{\delta}} < \frac{n^{a-\delta}}{a}.
$$
Taking any $\varepsilon > 0$, we have $n^{a-\delta}/a < \varepsilon$ if $n > (a\varepsilon)^{a-\delta}$, so taking $N:= \lceil (a\varepsilon)^{a-\delta} \rceil$ suffices to ensure the convergence.
A: Without L'Hospital: it is enough to show $\;\log\log n-\delta\log n\to-\infty$ in order to prove $\log n < n^\delta$. Now
$$\log\log n-\delta\log n=\log n\Bigl(\frac{\log\log n}{\log n}-\delta\Bigr)$$
Setting $y=\log n$, we know $\dfrac{\log y}y\xrightarrow[y\to\infty]{} 0$, hence the second factor tends to $-\delta$ as $n$ tends to $\infty$, and the product tends to $-\infty$ as asserted.
Added: Why $\displaystyle\lim_{y\to\infty} \dfrac{\ln y}y=0$
We suppose $y>0$. Start from $\;\dfrac1t<\dfrac1{\sqrt t}$ if $\;t>1$. By the mean value inequality we have
$$ \int_1^y \frac{\mathrm d\mkern1mu t}t=\ln y\le \int_1^y\frac{\mathrm d\mkern1mu t}{\sqrt t}=2\sqrt y-2<2\sqrt y$$
We deduce:
$$0<\frac{\ln y}y<\frac2{\sqrt y}\xrightarrow[y\to\infty]{} 0$$
A: One approach is to define the logarithm function as
$$\log x=\int_1^x\frac{1}{t}\,dt \tag 1$$
for $x>0$.  Then, we note that for any $\alpha<1$ and $x\ge1$, we have from $(1)$
$$\begin{align}
\log x&\le\int_1^x\frac{1}{t^{\alpha}}\,dt\\\\ &=\frac{x^{1-\alpha}-1}{1-\alpha}\tag 2
\end{align}$$
For any $\delta>1-\alpha$ we have from $(2)$
$$\frac{\log x}{x^{\delta}}\le \frac{x^{1-\alpha-\delta}-x^{-\delta}}{1-\alpha}\to 0 \,\,\text{as}\,\,x\to \infty$$
Therefore, for any $\epsilon>$, there is a number $N$ such that whenever $n>N$, 
$$\frac{\log n}{n^{\delta}}<\epsilon$$
Taking $\epsilon =1$, then $\frac{\log n}{n^{\delta}}<1$ whenever $n>N$.  And we are done!
