Show that ln x grows slower for $x \rightarrow \infty$ than every positive $x^α$

Question

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?

Answer 1

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!

Answer 2

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.

Answer 3

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 $.

Answer 4

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.