Proof ln(x) limit Goal: Prove that $ \lim_{x \to 0}\ln(x) = -\infty$.     
Consider $x \in (0,1]$ 
Fix $n \in \mathbb{N}$ 
Examine $\ln(\frac{1}{2^{n}}) = \int\frac{1}{t} \leq U(\frac{1}{t},P)$ 
Where our partition $P$ is $P =${$\frac{1}{2^{n}},...,\frac{1}{8},\frac{1}{4},\frac{1}{2},1$}    
Then $U(f(t),P) = \sum_{i=1}^{n}(\frac{1}{2^{i}}-\frac{1}{2^{i-1}})*2^{i-1}=-n\geq \ln(\frac{1}{2^{n}})\forall n \in \mathbb{N} $ 
Now fix $N \in \mathbb{N}$ and let $M $ be $M = \frac{1}{2^{2N+2}}$. We have that ln is strictly increasing which necessitates that $x < M$ and then $\ln(x) < \ln(M) = \ln(\frac{1}{2^{2N+1}}) \leq \ln(2^{-2(N+1)})=\frac{-2(N+1)}{2}=-N-1<-N $ 
Thus we can make $\ln(x)$ less than any $-N \in \mathbb{N}$ as $x \to 0$.   
I wrote this similarly to a proof we did in class and I guess I'm wondering why we have free choice of $M$ is it just like a $\delta$ in this proof or no? I'm a bit unsure about this proof.

$\ln x$ is an  increasing function of $x$ for $x>0$  (see Footnote) so it suffices to show that $\ln (2^{-n})\to -\infty$ as natural number $n\to \infty$ (see Remark).
For $n\in \Bbb N$ we have $$(\bullet)\quad \ln (2^{-n})=\int_1^{2^{-n}}(1/t)dt=\sum_{j=0}^{n-1}\int_{2^{-j}}^{2^{-j-1}}(1/t)dt.$$ Now $2^{-j}<2^{-j-1},$ and when $t\in [2^{-j-1},2^{-j})$ we have $(1/t)>2^j.$ So $$\int_{2^{-j}}^{2^{-j-1}}(1/t)dt<\int_{2^{-j}}^{2^{-j-1}}(2^j)dt=$$ $$=(2^{-j-1}-2^{-j})(2^j)=-1/2.$$ Applying this to $(\bullet)$ we have $$\ln (2^{-n})<\sum_{j=0}^{n-1}(-1/2)=-n/2.$$
Footnote. If $0<x<y$ then $\int_x^y(1/t)dt>0 $, so $$\ln y=\int_1^y(1/t)dt=\int_1^x(1/t)dt+\int_x^y(1/t)dt>\int_1^x(1/t)dt=\ln x.$$
Remark. For $0<x<1$ let $V(x)$ be the largest $n\in \Bbb N\cup \{0\}$ such that $x\le2^{-n}$. Then $V(x)\to \infty$ as $x\to 0,$ and $\ln  x\le \ln (2^{-V(x)})\le -V(x)/2.$
