Show $(\ln(x))^a<\sqrt x$ for any $a>1$ How do I show that for any $a>1$ there is a $C>0$ so that for any $x>C$ the inequation below is true:
$$ (\ln(x))^a<\sqrt x$$
I know that for a certain $a>1$ I can use the function:
  $$ f(x)=(\ln(x))^a-\sqrt x$$
And now to show that this function is monotonic (increasing) for a certain $M>0$. For a small $a$ I managed to show that, but for larger ones it becomes a hard work.
Moreover, I thought about showing that a limit of $$\lim_{x\to\infty} \frac {(\ln(x))^a}{\sqrt x}=0$$ 
which is known and then use the $\varepsilon$ definition of limit to show the inequation, but I don't know how to prove the limit above for any $a>0$ without using l'Hôpital a few times.
 A: Since $x>0$, you can set $x=t^{2a}$ with $t>0$; the inequality becomes then
$$
(\ln(t^{2a}))^a<\sqrt{t^{2a}}
$$
that is, since $\sqrt{t^{2a}}=t^a$,
$$
2a\ln t<t
$$
Consider
$$
f(t)=t-2a\ln t
$$
so that
$$
f'(t)=\frac{t-2a}{t}
$$
and the fact that
$$
\lim_{t\to\infty}f(t)=\infty
$$
A: Your idea is good. Given that
$$
\lim_{x\to\infty}\frac{\log(x)^a}{\sqrt{x}}=0\tag{1}
$$
and using $\epsilon=1$, we get that there is an $N$ so that if $x\ge N$, we have
$$
\left|\,\frac{\log(x)^a}{\sqrt{x}}-0\,\right|\lt1\tag{2}
$$
Using $(2)$, your inequality is straightforward.
To show $(1)$,
$$
\begin{align}
\lim_{x\to\infty}\frac{\log(x)^a}{\sqrt{x}}
&=\lim_{x\to\infty}\left(\frac{\log(x)}{x^{\frac1{2a}}}\right)^a\\[6pt]
&=\left(\lim_{x\to\infty}\frac{\log(x)}{x^{\frac1{2a}}}\right)^a\\
&=\left(\lim_{x\to\infty}\frac{\frac1x}{\frac1{2a}x^{\frac1{2a}-1}}\right)^a\\[6pt]
&=\left(\lim_{x\to\infty}2ax^{-\frac1{2a}}\right)^a\\[12pt]
&=0\tag{3}
\end{align}
$$
A: You can use l'Hopital to prove that 
$\lim \limits_{x \rightarrow \infty} \dfrac{ln(x)}{x^p} = 0$ for all $p > 0$. 
Then put $p = \dfrac{1}{2a}$, use $\varepsilon$ argument to show that $ln(x) < x^{\frac{1}{2a}}$ and raise both to power $a$. You don't need to worry about reversing the inequality or getting undefined numbers because we are letting $x$ go to $\infty$.
A: 1. This answer tries to delineate what is considered to be known and not known to prove the desired result. We only assume that one knows the limit 

$$\lim_{x\to\infty}\frac{\log x}x=0\tag{$\ast$}$$ 

2. Then one also knows that, for every positive $(\beta,\gamma)$, $$\lim_{x\to\infty}\left(\frac{\log(x^\beta)}{x^\beta}\right)^\gamma=0\tag{$\ast\ast$}$$ Since $$\left(\frac{\log(x^\beta)}{x^\beta}\right)^\gamma=\left(\beta\frac{\log x}{x^\beta}\right)^\gamma=\beta^\gamma\frac{(\log x)^\gamma}{x^{\beta\gamma}}$$ the choice $$(\beta,\gamma)=(1/(2a),a)$$ in $(\ast\ast)$, for any positive $a$, yields

$$\lim_{x\to\infty}\frac{(\log x)^a}{\sqrt x}=0$$ 

3. Likewise, $(\ast)$ implies that, for every positive $(a,b)$,$$\lim_{x\to\infty}\frac{(\log x)^a}{x^b}=0$$
4. Finally, to show $(\ast)$, note that for every $x>1$ and every $z$ in $(1,x)$, $$\log x=\int_1^x\frac{dt}t=\log z+\int_z^x\frac{dt}t\leqslant\log z+\int_z^x\frac{dt}z=\log z+\frac{x}z-1$$ hence, for every fixed $z>1$, $$\limsup_{x\to\infty}\frac{\log x}x\leqslant\limsup_{x\to\infty}\left(\frac{\log z-1}x+\frac1z\right)=\frac1z$$ This holds for every $z>1$ hence $(\ast)$ holds.
A: Hint: Compute the limit
$$\lim_{x\to\infty}\frac{\ln x}{\ln\ln x}$$
to show that for some $C>0$, $x>C$ implies $\ln x>2a\ln\ln x$.
A: The fundamental inequality satisfied by $\log$ function is $$\log t\leq t - 1,\forall t>0\tag{1}$$ and equality holds only for $t=1$ (you should be able to prove this yourself using any suitable definition of $\log $ function without any difficulty). Here we use the inequality with $t>1$. Let's put $t=x^{b} $ where $b>0$ will be determined (later) based on value of $a$. Then we can see that the above inequality is transformed into $$b\log x<x^{b} - 1<x^{b}$$ or $$(\log x) ^{a} <\frac{x^{ab}} {b^{a}}\tag{2} $$ Next we choose the constant $b$ such that $ab<1/2$ so that $$\frac{x^{ab} /b^{a}} {\sqrt{x}} \to 0$$ as $x\to\infty$. It follows that there is a $C>1$ such that $(x^{ab} /b^{a}) /\sqrt{x} <1$ for all $x>C$. And thus from $(2)$ we arrive at $(\log x) ^{a} <\sqrt{x} $ for all $x>C$.

Note that the result holds for all $a>0$ which includes the given condition $a>1$. It is interesting to note that the simple looking inequality $(1)$ is the key behind the well known limits $$\lim_{x\to\infty} \frac{\log x} {x} =0,\lim_{x\to\infty}\frac{(\log x) ^{a}} {x^{b}} =0,\forall a>0,b>0\tag{3}$$ and the above two limits are equivalent (given one the other can be proved via algebraic manipulation). The power of inequality $(1)$ is enhanced greatly by the aid of the properties of $\log$ function like $b\log x=\log x^{b} $ which is used in above proof. 
