What am I missing in this argument for $\lim\limits_{x\rightarrow \infty} \ln x = \infty$? In an appendix of Stewart's Calculus, the logarithmic and exponential functions are built up starting from the defnition $\ln x = \int_1^x \frac{1}{t}\,dt$.
Having shown that $\ln(x^n) = n \ln(x)$ for integer $n$, the following argument is then made for $\lim\limits_{x\rightarrow \infty} \ln x = \infty$:

With $x=2$ and n any positive integer, we have $\ln(2^n) = n\ln 2$.
  Now $\ln 2 > 0$, so this shows that $\ln(2^n) \rightarrow \infty$ as
  $n\rightarrow \infty$.
But $\ln x$ is an increasing function since its derivative
  $\frac{1}{x} > 0$. Therefore $\ln x \rightarrow \infty$ as $x \rightarrow \infty$.

The first sentence makes perfect sense.
The second sentence, taken on its own, is a false argument (e.g. $1-e^{-x}$ has a positive derivative, yet it goes to $1$ not $\infty$).
I can show by other means that $\lim\limits_{x\rightarrow \infty} \ln x = \infty$. My question is what am I missing in this specific argument, which at the moment seems to me like a non sequitur.
 A: It is meant to connect $\lim_{n \to \infty}\ln(2^n)$, which is just saying something about what $\ln$ is doing on very specific, discrete points, with $\lim_{x \to \infty} \ln (x)$ which says what happens to the function at all points.
For contrast, take
$$
f(x) = x\cos(2\pi x)
$$
where we have
$$
\lim_{n \to \infty} f(n) = \lim_{n \to \infty}n = \infty
$$
but the function is far from monotonous, so we cannot conclude that $\lim_{x \to \infty} f(x) = \infty$.
A: You know that
$$\lim_{x\to+\infty} f(x)=+\infty\stackrel{\operatorname{def}}{\iff} \forall M>0,\, \exists \alpha_M>0\ \forall x\ge\alpha_M,\, f(x)\ge M$$
You have already shown that $$\forall M>0,\,\exists n_M\ \forall m\ge n_M,\ \ln(2^m)\ge M$$
Expecially, $\ln(2^{n_M})\ge M$.
Now, since $\ln$ is a growing function (since its derivative is strictly positive), if you pick $\alpha_M=2^{n_M}$ you know that $$x\ge \alpha_M\implies \ln(x)\ge \ln(\alpha_M)=\ln(2^{n_M})\ge M$$
Whence the thesis for $\alpha_M$ defined as above.
A: He uses the monotonicity to make a comparison. For $x>2$ by monotonicity 
$$\ln(x^n)>\ln(2^n) \longrightarrow +\infty$$ as $n \to \infty$ (by his first argument). So, the LHS goes also to $\infty$. In your example with $1-e^{-x}$ this comparison in not possible because the RHS goes to $0$. 
