# How to evaluate $\lim\limits_{s\to\infty}\log s$?

I am stuck with applying limit at the following step, limit $$\lim\limits_{s\to\infty}\log s.$$ Now I am unable to do anymore steps(I cant figure out how do I apply the limit and get a valid answer). Please help me out.

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There's no denominator. Anyway, what definition of $\ln$ are you using? If the integral definition, you're asking what the integral $\int_1^\infty \frac{\mathrm dt}{t}$ is... –  Ｊ. Ｍ. Oct 31 '11 at 16:45
I tried to guess which limit did you mean. It it right? –  Ilya Oct 31 '11 at 16:46
@akito - I'm not sure what you are asking. Do you want the limit of ln(s) as s becomes infinite? You have both s and S in your question. Are they the same thing, or is S dependent on s in some way? If they are the same, then it should be straightforward to show that ln(s) approaches infinity as s does. Otherwise, the question needs some clarifying. –  Chris Leary Oct 31 '11 at 16:47
@Akito: What is the base of your $log$? –  Hassan Muhammad Oct 31 '11 at 16:54
@Hassan: does it matter here? –  Ｊ. Ｍ. Oct 31 '11 at 16:55

The statement $$\lim_{x\to\infty}\log x=+\infty$$

is true if and only if for each $M>0$ there exists $N>0$ such that whenever $x>N$ we have $\log x>M$. Given $M$, take $N=e^M$. Since the logarithm is an increasing function, whenever $x>N=e^M$ we will have $\log x>\log N=M$, thus $$\lim_{x\to\infty}\log x=+\infty$$

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Shouldn't those limits be equal to infinite instead of tending towards infinite? Note: I'm not saying there aren't limits that tend towards something, I'm saying these don't. Perhaps it was just a distraction while you were writing, I comment on this for a matter of precision. –  JMCF125 Jun 7 '13 at 18:36
@JMCF125 It is sometimes a mater of taste, but thank you. –  Pedro Tamaroff Jun 7 '13 at 21:57

\begin{align} \log_2 2 & = 1 \\ \log_2 4 & = 2 \\ \log_2 8 & = 3 \\ \log_2 16 & = 4 \\ \log_2 32 & = 5 \\ & \vdots \end{align} As the number on the left (2, 4, 8, 16, 32,...) approaches $\infty$, so does the one on the right.

Similarly with other bases than 2, as long as the base is more than $1$.

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@Michel Hardy: This is a particular case, can you prove it generally? Let us see the argument clearly for every real base $\gt1$ and every real number $s\gt1$. –  Hassan Muhammad Oct 31 '11 at 17:51
@Hassan $\log_b s = \frac{\log_2 s}{\log_2 b}$, so the limit is $\infty$ for any base $b > 1$... –  Srivatsan Oct 31 '11 at 18:07
As I said: Similarly for other bases greater than $1$. –  Michael Hardy Nov 1 '11 at 4:25
@MichaelHardy: This is, alas, the way of the world. No, I did not downvote, but I find your presentation above to be far from a proof. Looking at a few terms of an infinite sequence is never enough, by itself, to determine the limit. Worse when you look at the first few terms of a subsequence... Yes, you could expand and make it correct. However, you didn't. –  Arturo Magidin Nov 2 '11 at 4:31
@Arturo: I don't see any claim of a proof in the answer, which is not surprising when the question did not mention proof. Moreover, there is no "looking at a few terms ... to determine the limit", and no expansions needed to make the answer correct. It is a sketch, brief, correct and quite a good hint on how to write down a proof. –  zyx Nov 2 '11 at 6:36

The logarithm is alike the square root, it tends towards infinity as $x$ does too. The other guys already said that, I'll attempt to prove it. We know that (and that for any log, not just the natural one, I'm using this one because it requires a smaller amount of letters, although that is probably lost with this explanation)

$$\frac{\ln(x)}{\ln(y)}=\log_y(x)$$

so

$$\lim_{x \to +\infty}\log_b(x) = \lim_{x \to +\infty}\frac{\ln(x)}{\ln(b)} = \lim_{x\to +\infty}\ln(x) = \lim_{x\to + \infty}\left( \ln \left( \frac x {e^m}\right)+m \right): m\in \mathbb N, \\ \frac x {e^m} \not\to 0,\frac x {e^m}\not\to\infty\\ \therefore \lim_{x\to + \infty}\big(m\big) =\infty$$ As $m$ is just big/small enough for the above to hold (that is, $\frac x {e^m}\not\to 0, \frac x {e^m}\not\to \infty$), the value of the logarithm can be ignored, because it will thus not become $-\infty$ (this is probably the least formal part of this attempt of proof, but I think it still conforms to the definition of $m$) $$\lim_{x\to + \infty}\left( \ln \left( \frac x {e^m}\right)+m \right) = \lim_{x\to + \infty}\big(m\big) =\infty$$

as $$\forall a \in \mathbb{R}, \forall b \in\mathbb R,\lim_{x\to \infty} (ax + b) = \lim_{x\to \infty} (x) = \infty$$

because $a$ and $b$ are constants (as above, with constants $\frac 1 {\ln(b)}$ and bounded variable $\ln \left( \frac x {e^m}\right)$).

## Conclusion

$$\forall b>1,\ \lim_{x\to+\infty}\log_b(x)=+\infty\\ \forall b:0<b<1,\ \lim_{x\to+\infty}\log_b(x)=-\infty$$ The first one is obvious from what is above, the second arises from the fact $\frac x {e^m}$ is bounded even when $m\to-\infty$.

I hope this was an actual proof, not just a conjecture, and that it is correct and clear for understanding.

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If you observe graph below which represents $\log(s)$ you may conclude that:

$\lim\limits_{s\to+\infty}\log s=+\infty$

$\lim\limits_{s\to-\infty}\log s$ isn't defined since $s>0$

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How do you conclude that $\log s \to \infty$ if the graph only goes up to 10? –  Pedro Tamaroff Feb 8 '12 at 20:39