# Prove that $\lim_{x\to\infty} (\ln x) = \infty$

Can someone help me prove that the function $\ln(x)$ diverges to infinity as $x$ approaches infinity. I tried using the definition to show that $\lvert \ln(x) -∞ \rvert < \epsilon$ where $\epsilon > 0$ and there exists a number $N$ which is in the set of natural numbers where $x > N$, proving that it does indeed diverge to infinity but I can't get any further.

Any help is appreaciated.

• $\ln$ is monotonically increasing. So, you only have to show that $\ln$ is not bounded. This follows from $\ln(e^x) = x$. – Friedrich Philipp Apr 2 '17 at 22:47
• Looking at your question history ($5$ closed questions out of your $7$ for a total of $-12$ votes) I don't even know why I bother but anyway, may I point out to you what you already know: This post does not match users' quality standards, so it will attract downvotes and/or be put on hold. For information about writing math on MSE see here, here, here and here. – user409521 Apr 2 '17 at 22:49
• You should look again at the definition of “diverges to infinity”: $\ln(x)-\infty$ means nothing at all. – egreg Apr 2 '17 at 23:09
• @egreg: $\ln(x) - \infty$ means $-\infty$ so long as $x \neq +\infty$. – Hurkyl Apr 2 '17 at 23:12
• @Hurkyl No, not in that context, at least. – egreg Apr 2 '17 at 23:23

Hint:

For all $t\ge 1$, one has: $$0 <\sqrt t\le t, \enspace\text{hence}\quad \frac1t\ge\frac1{\sqrt t},\enspace\text{so}\quad \int_1^x\frac{\mathrm dt}{t}\ge \int_1^x\frac{\mathrm dt}{\sqrt t}.$$

• Wouldn't the fact that $0 < \sqrt{t} \le t$ show that $1/t \le 1/\sqrt{t}$? – 2.71828-asy Apr 2 '17 at 22:53

Your error is that you have the wrong definition of limit.

Finite limit (with value $L$):

$\forall \epsilon > 0 :\; \cdots \implies |f(x) - L| < \epsilon$

Limit is $+\infty$

$\forall M:\; \cdots \implies f(x) > M$

Limit is $-\infty$

$\forall M:\;\cdots \implies f(x) < M$

You used the version for finite limits, rather than the version for a limit equal to $+\infty$.

Hint:

show that if $x > \lceil\exp(M+1) \rceil$, then we must have

$$\ln(x)>M$$

I guess you want an analysis using the definition of divergence at infinity.

Let $M > 0$; and note that $\log x > M$ if $x > e^{M}$. This shows that for every $M > 0$ there is some $X > 0$ (take $X := e^{M}$, say) such that $x > X$ implies $\log x > M$.