$\lim_{x\to \infty} \ln x=\infty$ I'm reading the following reasoning: 

Since $\underset{n\to \infty}{\lim}\ln 2^n=\underset{n \to \infty}{\lim}n\cdot(\ln 2)=\infty$ then necessarily $\underset{x\to \infty}{\lim}\ln x =\infty$.

I don't understand how the generalisation was done from $\lim_{n\to \infty}\ln 2^n=\infty$ to $\lim_{x\to \infty} \ln x=\infty$. 
Thank you for your help!
 A: The logarithm is monotone, so it is enough to look at what happens at the integers. Or even the integers of the form $2^n$. 
More concretely, given $K>0$, there exists $n_0$ with $\ln 2^{n_0}\geq K$. Then, if $x\geq 2^{n_0}$, we have 
$$
\ln x\geq\ln 2^{n_0}\geq K. 
$$
So $\lim_{x\to\infty}\ln x=\infty$. 
A: The idea is that we can make $\ln x$ arbitrarily large by making $x$ sufficiently large. Hence, say we want to make $\ln x > c$ for arbitrarily large $c$. Then for some $n>\frac{c}{\ln 2}$ take 
$x>2^n$ so that $\ln x > n \ln 2 > c$ noting that $\ln$ is increasing.  
A: In order to show that $\lim_{x \to \infty}\log x = \infty$ we need to show that for every $M > 0$ there is an $N > 0$ such that $\log x > M$ for all $x > M$. And you already know that $\lim_{n \to \infty}\log 2^{n} = \infty$ so that given any $M_{1} > 0$ there is a positive integer $m$ such that $\log 2^{n} > M_{1}$ for all $n \geq m$.
Let's see how we can achieve our goal from the given information. First idea is to take $M_{1} = M$ so that there is an integer $m > 0$ such that $\log 2^{n} > M$ for all $n \geq m$. Now we know that if $x > 2^{n} \geq 2^{m}$ then $\log x > \log 2^{n} > M$ and thus it is sufficient to take $N = 2^{m}$ and you are done.
The fundamental idea is that $\log x$ is monotone (we don't need strictly monotone nature in the above proof) and hence in order to prove that $\log x \to \infty$ as $x \to \infty$ it is sufficient to consider a sequence $s_{n}$ of positive numbers such that $s_{n} \to \infty$ as $n \to \infty$ and show that $\log s_{n} \to \infty$. Here we have taken $s_{n} = 2^{n}$ and we could have taken $s_{n} = a^{n}$ with any $a > 1$. The reason we take $s_{n}$ of the form $a^{n}$ is that we have the property $\log a^{n} = n\log a$ and this makes it easy to show that $\log s_{n} \to \infty$. If we take any other sequence say $s_{n} = 2n$ then we get $\log 2n = \log 2 + \log n$ then we need to establish that $\log n$ tends to infinity and this is back to the problem with which we started.
A: For $x,y > 0$, $\ln x > \ln y$ if and only if $x > y$.  Also, for any $n$ you can find an $x$ for which $x > 2^n$.
This shows for any $S > 0$ there exists an $x_0$ such that $\ln x > S$ for all $x > x_0$.  That is what it means for $\ln x$ to have limit $\infty$ as $x \to \infty$.
A: For $2\leq n\in N$ and  $x\geq n$ we have $$\ln x=\int_1^x(1/y)\;dy\geq \int_1^n(1/y)\;dy=\sum_{j=1}^{n-1}\int_j^{j+1}(1/y)\;dy\geq \sum_{j=1}^{n-1}\int_j^{j+1}(1/(j+1))\;dy=$$ $$=\sum_{j=1}^{n-1}1/(j+1).$$ There are many ways to show that this last sum has no upper bound as $n\to \infty.$ The simplest, I think,is $$1/2+1/3+1/4\;>\;1/2+2(1/4)=1,$$  $$1/2+1/3+...1/8\;>\;1/2+2(1/4)+4(1/8)=3/2,$$  $$1/2+1/3+...+1/16\;>\;1/2+2(1/4)+4(1/8)+8(1/16)=2,$$...... et cetera. 
