$\lim\limits_{x \to \infty} \frac{\ln x}{x} =0$

Question

I wanted to prove $\lim\limits_{x \to \infty} \frac{\ln x}{x} =0$ by the squeeze theorem.

I know $$e^x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$$ by Taylor expansion, so $e^x \ge 1+x$ and thus $x \ge \ln(1+x)$ and finally, $x \ge \ln(x)$.

I also know $\ln x \ge 0$ for $x\geq 1$.

So I tried to apply the squeeze theorem $0 \le \ln x \le x$, and then dividing throughout by $x$ I get $0 \le \frac{\ln x}{x} \le 1$ and then if I apply limit $x \to \infty$, but I'm unable to apply the squeeze theorem here.

Can anybody help me understand what is it that I'm doing wrong? I do not want to use L'Hospital's theorem.

Answer 1

$\ln x=2 \ln x^{\frac 12}\le2\sqrt x$ (using the inequality $y\ge \ln y$ for $y\gt 0$ that you obtained.)

So it follows that: for all $x\gt 0$ $$|\frac{\ln x}x|\le2 \frac{\sqrt x}x$$

It follows by squeeze theorem that $\lim_{x\to \infty}|\frac{\ln x}x|=0.$ Now note that $-|\frac{\ln x}x|\le \frac{\ln x}x\le|\frac{\ln x}x|$ and hence the result follows by squeeze theorem.

Answer 2

I think you want to show: $\displaystyle \lim_{x \to \infty} \dfrac{\ln x}{x} = 0$. This is immediate since $0 < \dfrac{\ln x}{x} < \dfrac{1}{\sqrt{x}}, x > 10$.

Answer 3

Extending the approximation

$$e^x\ge 1+x$$

into

$$e^x\ge 1+x+\frac{x^2}{2}>\frac{(x+1)^2}{2}$$

and

$$e^x\ge 1+x+\frac{x^2}{2}+\frac{x^3}{6}>\frac{(x+1)^3}{6}$$

allows

$$e^x> \frac{(x+1)^n}{n!}$$

for all $n\in\mathbb{Z}$.

So,

$$x\ge n\ln(1+x)-\ln n!\ge n\ln x - \ln n!$$

So

$$\frac{x}{n}+\frac{\ln n!}{n}\ge \ln x$$

$$\frac{1}{n}+\frac{\ln n!}{nx}\ge \frac{\ln x}{x}$$

As we can make both terms on the LHS arbitrarily small, we are done.

Answer 4

Here's a way with the integral definition.

If $x > 1$ then

$\ln(x) =\int_1^x \dfrac{dt}{t} $ so, for any $c > 0$,

$\begin{array}\\ \ln(1+x) &=\int_1^{1+x} \dfrac{dt}{t}\\ &=\int_0^{x} \dfrac{dt}{1+t}\\ &<\int_0^{x} \dfrac{dt}{(1+t)^{1-c}} \qquad\text{(since } (1+t)^{1-c} < 1+c\\ &=\int_0^{x} (1+t)^{c-1}dt\\ &=\dfrac{(1+t)^c}{c}\big|_0^x\\ &=\dfrac{(1+x)^c-1}{c}\\ &<\dfrac{(1+x)^c}{c}\\ \text{Therefore}\\ \dfrac{\ln(1+x)}{(1+x)^{2c}} &<\dfrac1{c(1+x)^c}\\ \end{array} $

or, for $x > 1$ and any $c > 0$, $\dfrac{\ln(x)}{x^{2c}} \lt\dfrac1{cx^c} $.

For example, if $c = \frac12$, this gives $\dfrac{\ln(x)}{x} \lt\dfrac{2}{x^{1/2}} $.

Replacing $c$ by $c/2$, this gives $\dfrac{\ln(x)}{x^{c}} \lt\dfrac{2}{cx^{c/2}} $ which shows that $\dfrac{\ln(x)}{x^{c}} \to 0$ as $x \to \infty$ for any $c > 0$.