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

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

## marked as duplicate by rtybase, Robert Wolfe, Hans Lundmark, Simply Beautiful Art, Lord Shark the UnknownJan 20 at 5:33

• Hint: You can use the change of variable $x = e^y$. – Matheus Manzatto Jan 20 at 1:05
• Why you haven´t applied l´hospital? – callculus Jan 20 at 1:07
• because I can't use that there. – VirtualUser Jan 20 at 1:08
• Here, Propositions 2.2 is a proof of $\frac{x^{\varepsilon}}{\ln{x}} \rightarrow \infty$ – rtybase Jan 20 at 1:47
• Here is another one ... – rtybase Jan 20 at 1:54

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

Hint:

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

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

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