Vtgyjfy

I want to calculate $$limlimits_{ xto infty} frac{ln(x)}{x^a}$$ where $ a > 0 $

It looks simple because if $a>0$ then $ x^a $ it grows asymptotically faster than $ ln(x) $ so
$$limlimits_{ xto infty} frac{ln(x)}{x^a} = 0$$
But I don't know how to formally justify that. I am thinking about something what I was doing in case of sequences:
$$frac{ln(x+1)}{(x+1)^a} cdot frac{x^a}{ln(x)} $$
But it have no sense because sequences was being considered in $mathbb N$ but functions like that are considered in $mathbb R$
I can't use there hospital's rule

Lemma: Let $f,g: mathbb{R}to mathbb{R}$ be continuous functions such that $limlimits_{trightarrow infty} g(t) = infty$ and
$$limlimits_{trightarrow infty} fleft(g(t)right) = L, $$
then $$limlimits_{trightarrow infty} f(t) = L.$$

Proof: We need to show that for every $varepsilon>0$, there exists $M >0$, such that
$$forall t>M Rightarrow left|f(t) - Lright|<varepsilon. $$

Let $varepsilon$ be a number greater than zero, once $limlimits_{trightarrow infty} fleft(g(t)right) = L$, there exists $M_1 >0$ such that

$$forall t>M_1 Rightarrow |f(g(t)) - L|<varepsilon. quad (1) $$

Define $M_2 := inf{g(t), t > M_1}$. Once $g(t) rightarrow infty$ when $t rightarrow infty$, by the mean value theorem, for every $s> M_2$, there exists $z> M_1$ satisfying $g(z) = s$. Therefore using the previous conclusion and (1) we are able to conclude

$$forall s > M_2 Rightarrow |f(s) - L | < varepsilon, $$
which demonstrates the lemma.

Now define the functions ($a>0$) $$f(x) = frac{ln(x)}{x^a} text{and} g(x) = e^x.$$

Note that $$f(g(x)) = frac{x}{e^{ax}},$$ implying (because $a>0$)

$$lim_{x rightarrow infty} frac{x}{e^{ax}} = 0,$$
and using our lemma, we conclude that
$$lim_{t rightarrow infty} frac{ln(t)}{t^a} = 0. $$

Hint:

For $x > 1$ and $0 < b < a$,

$$0 < frac{ln x}{x^a} = frac{1}{b} frac{ln x^b}{x^a} < frac{x^b}{bx^a}$$

$$
x^a=e^{aln x}=1+aln x+frac12 (aln x)^2+cdots
$$
Hence for large $x>1$,
$$
0lefrac{ln x}{x^a}lefrac{ln x}{1+aln x+frac12 (aln x)^2}=
frac{1}{frac{1}{ln x}+a+frac12 a^2ln x}.
$$
But
$$
frac{1}{frac{1}{ln x}+a+frac12 a^2ln x}to 0
$$
as $xto infty$.