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.
 A: 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 $$\lim_{x\to\infty}\log x=+\infty$$
A: $$
\begin{align}
\log_2 2 & = 1 \\
\log_2 4 & = 2 \\
\log_2 8 & = 3 \\
\log_2 16 & = 4 \\
\log_2 32 & = 5 \\
& \vdots
\end{align}
$$
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 $1$.
A: 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.
A: 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 $s>0$

