Calculate $\lim\limits_{ x\to \infty} \frac{\ln(x)}{x^a}$ where $ a > 0 $

Question

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

Answer 1

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.

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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 $$\lim_{t \rightarrow \infty} \frac{\ln(t)}{t^a} = 0. $$

Answer 2

Hint:

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

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

Answer 3

$$ x^a=e^{a\ln x}=1+a\ln x+\frac12 (a\ln x)^2+\cdots $$ 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 $x\to \infty$.