$\lim\limits_{x \to \infty} \frac{\ln x}{x} =0$ 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.
 A: $\ln x=2 \ln x^{\frac 12}\le2\sqrt x$ (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.
A: I think you want to show: $\displaystyle \lim_{x \to \infty} \dfrac{\ln x}{x} = 0$. This is immediate since $0 < \dfrac{\ln x}{x} < \dfrac{1}{\sqrt{x}}, x > 10$.
A: 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.
A: 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 $c > 0$.
