$\lim_{x\to \infty} \ln x=\infty$ 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!

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. 
