Show that ln x grows slower for $x \rightarrow \infty$ than every positive $x^α$ I have to prove that the logarithmus
$\ln x$ grows slower for $x \rightarrow \infty$ than every positive $x^α$ ($α>0$).
This is my approach:
If $\displaystyle\lim\limits_{x \rightarrow \infty}{ \frac{f(x)}{g(x)}} = 0$ then $f(x)$ grows slower than $g(x)$.
Let $\displaystyle f(x) = \ln{x}$ and $\displaystyle g(x)=x^α$.
$\displaystyle \lim\limits_{x \rightarrow \infty}{ \frac{\ln{x}}{x^α}} = \frac{∞}{∞} \Rightarrow
\lim\limits_{x \rightarrow \infty}{ \frac{\frac{1}{x}}{α \cdot x^{α-1}}} = \lim\limits_{x \rightarrow \infty}{ \frac{1}{x \cdot α \cdot x^{α-1}}} = 0$
By applying L'Hospital's rule, we can see that $f(x)$ grows slower than $g(x)$. Am I right?
 A: This is correct. However, saying that 
$$\lim_{x\to\infty} \frac{\ln x}{x^{\alpha}} = \frac{\infty}{\infty}$$
is not great, since $\frac{\infty}{\infty}$ isn't properly defined. In addition, you should probably explain why 
$$\lim_{x\to\infty} \frac{1}{x\alpha x^{\alpha-1}}$$
converges to $0$ as opposed to just stating that it does. Finally, you should use double dollar signs for separate line equations to make them look nicer; compare
$\lim_{x\to\infty} \frac{x}{x} = 1$
with
$$\lim_{x\to\infty} \frac{x}{x} = 1$$
Other than that, the proof looks good!
A: To prove that
$$\lim_{x\to+\infty}\frac{\ln(x)}{x^\alpha}=0,$$
taking logarithm, we get
$$\lim_{x\to+\infty}\ln(x)\left( \frac{\ln(\ln(x))}{\ln(x)}-\alpha \right)=-\infty$$
since $$\lim_{X\to+\infty}\frac{\ln(X)}{X}=0$$
qed.
A: Here is a proof that
$\lim_{x \to \infty} \dfrac{\ln x}{x}
=0
$
is all you need.
Suppose that
$\lim_{x \to \infty} \dfrac{\ln x}{x}
=0
$.
Then,
for any $a > 0$,
$\lim_{x \to \infty} \dfrac{\ln x^a}{x^a}
=0
$.
Therefore
$\lim_{x \to \infty} \dfrac{\ln x}{x^a}
=\lim_{x \to \infty} \dfrac1{a}\dfrac{a\ln x}{x^a}
=\dfrac1{a}\lim_{x \to \infty} \dfrac{\ln x^a}{x^a}
=0
$.
A: Let $f(x)=\dfrac{\ln x}{x^{\alpha}}$, $x\geq 1$, then $f'(x)=\dfrac{x^{\alpha}x^{-1}-(\ln x)\alpha x^{\alpha-1}}{x^{2\alpha}}=\dfrac{x^{\alpha-1}(1-\alpha\ln x)}{x^{2\alpha}}$. Let $f'(x)=0$, then $x=e^{1/\alpha}$, but it is easy to see that $f'(x)<0$ for $x>e^{1/\alpha}$ and that $f'(x)>0$ for $x<e^{1/\alpha}$, so $f(x)\leq f(e^{1/\alpha})=\dfrac{1}{\alpha e}$, so $\ln x\leq\dfrac{1}{\alpha e}x^{\alpha}$ for all $x\geq 1$. We conclude that we find some constant $C_{\alpha}$ which depends only on $\alpha$ such that $\ln x\leq C_{\alpha}x^{\alpha}$ for $x\geq 1$, so there is some constant $C_{\alpha/2}$ such that $\ln x\leq C_{\alpha/2}x^{\alpha/2}$ for all such $x$, hence $\dfrac{\ln x}{x^{\alpha}}\leq C_{\alpha/2}\dfrac{1}{x^{\alpha/2}}$, taking $x\rightarrow\infty$ will give you the result.
