# What is the limit as x approaches infinity of ln(x)?

Mar 24, 2015

${\lim}_{x \rightarrow \infty} \ln x = \infty$

To see this, we'll use:

$\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$

and

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} = {\int}_{a}^{c} f \left(t\right) \mathrm{dt} + {\int}_{c}^{b} f \left(t\right) \mathrm{dt}$

and

If, on $\left[a , b\right]$ we have $f \left(t\right) \ge m$, then ${\int}_{a}^{b} f \left(t\right) \mathrm{dt} \ge \left(b - a\right) \cdot m$

We will look at intervals of the form: $\left[{2}^{n} , {2}^{n + 1}\right]$

On $\left[1 , 2\right]$, we have $\frac{1}{t} \ge \frac{1}{2}$, so
${\int}_{1}^{2} \frac{1}{t} \mathrm{dt} \ge \left(2 - 1\right) \cdot \frac{1}{2} = \frac{1}{2}$
And so, $\ln 2 \ge \frac{1}{2}$

On $\left[2 , 4\right]$, we have $\frac{1}{t} \ge \frac{1}{4}$, so
${\int}_{1}^{4} \frac{1}{t} \mathrm{dt} = {\int}_{1}^{2} \frac{1}{t} \mathrm{dt} + {\int}_{2}^{4} \frac{1}{t} \mathrm{dt} \ge \frac{1}{2} + \left(4 - 2\right) \cdot \frac{1}{4} = \frac{1}{2} + \frac{1}{2} = 1$
And, $\ln 4 \ge \frac{2}{2} = 1$

.

On each $\left[{2}^{n} , {2}^{n + 1}\right]$, we have 1/t >= 1/(2^(n+1) So the additional integral ${\int}_{{2}^{n}}^{{2}^{n + 1}} \frac{1}{t} \mathrm{dt}$ adds more than

$\left({2}^{n + 1} - {2}^{n}\right) \cdot \frac{1}{{2}^{n + 1}} = \left[{2}^{n} \left(2 - 1\right)\right] \cdot \frac{1}{2} ^ \left(n + 1\right) = \frac{{2}^{n}}{2} ^ \left(n + 1\right) = \frac{1}{2}$

And so, $\ln \left({2}^{n + 1}\right) = {\int}_{1}^{{2}^{n + 1}} \frac{1}{t} \mathrm{dt} \ge \frac{n + 1}{2}$

So, as $x \rightarrow \infty$, we have ${\int}_{1}^{x} \frac{1}{t} \mathrm{dt} \rightarrow \infty$.

Since this integral is $\ln x$, we have ${\lim}_{x \rightarrow \infty} \ln x = \infty$

May 3, 2015

The answer is $+ \infty$

You can prove it by reductio ad absurdum.

You know that if $x > 1 \ln \left(x\right) > 0$ so the limit must be positive.
You also know that $\ln \left({x}_{2}\right) - \ln \left({x}_{1}\right) = \ln \left({x}_{2} / {x}_{1}\right)$ so if ${x}_{2} > {x}_{1}$ the difference is positive, so $\ln \left(x\right)$ is always growing

If ${\lim}_{x \to \infty} \ln \left(x\right) = M \in \mathbb{R}$ you have $\ln \left(x\right) < M \implies x < {e}^{M}$, but $x \to \infty$ so $M$ can not be in $\mathbb{R}$, and the limit must be $+ \infty$

Jul 28, 2017

For any strictly increasing function, f, if $a < b$, then $f \left(a\right) < f \left(b\right)$.
Let M be an arbitrary positive number.
Since (as others have shown), $f \left(x\right) = \ln x$ is strictly increasing on $\left(0 , \infty\right)$,
$\ln x > M$
if and only if
$x > {e}^{M}$.

Since the numbers themselves increase without bound, we have shown that by making x large enough, we may make $f \left(x\right) = \ln x$ as large as desired.

Thus, the limit is infinite as x goes to $\infty$.