Calculate $\lim\limits_{ x\to \infty} \frac{\ln(x)}{x^a}$ where $ a > 0 $ I want to calculate $$\lim\limits_{  x\to \infty} \frac{\ln(x)}{x^a}$$ where $ a > 0 $ 
It looks simple because if $a>0$ then $ x^a $ it grows asymptotically faster than $ \ln(x) $ so
$$\lim\limits_{  x\to  \infty} \frac{\ln(x)}{x^a} = 0$$
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
 A: Hint:
For $x > 1$ and $0 < b < a$,
$$0 < \frac{\ln x}{x^a} = \frac{1}{b} \frac{\ln x^b}{x^a} < \frac{x^b}{bx^a}$$
A: 
Lemma: Let $f,g: \mathbb{R}\to \mathbb{R}$ be continuous functions such that $\lim\limits_{t\rightarrow \infty} g(t) = \infty$ and 
  $$\lim\limits_{t\rightarrow \infty} f\left(g(t)\right) = L, $$
  then $$\lim\limits_{t\rightarrow \infty} f(t) = L.$$

Proof: We need to show that for every $\varepsilon>0$, there exists $M >0$, such that 
$$\forall \ t>M \Rightarrow \ \left|f(t) - L\right|<\varepsilon. $$
Let $\varepsilon$ be a number greater than zero, once $\lim\limits_{t\rightarrow \infty} f\left(g(t)\right) = L$, 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
$$\lim_{t \rightarrow \infty} \frac{\ln(t)}{t^a} = 0. $$
A: $$
x^a=e^{a\ln x}=1+a\ln x+\frac12 (a\ln x)^2+\cdots
$$
Hence for large $x>1$,
$$
0\le\frac{\ln x}{x^a}\le\frac{\ln x}{1+a\ln x+\frac12 (a\ln x)^2}=
\frac{1}{\frac{1}{\ln x}+a+\frac12 a^2\ln x}. 
$$
But 
$$
\frac{1}{\frac{1}{\ln x}+a+\frac12 a^2\ln x}\to 0
$$
as $x\to \infty$.
