Why does $\lim_{x\to 0^{+}} \ln(e^x - 1) = -\infty$ and not $1$? What does it mean for a limit to be approaching from the right? I know the graph of $\ln(x)$, and I understand that if $\ln(x)$ is being approached from the righthand side that means it's heading towards $-\infty$ but wouldn't it be like this:
$\lim_{x\to 0^+}\ln(e^{0^+}-1)$
I don't know what $e^{0^+}$ actually represents I think it would just be $e^0$ in that case, right?
So then,
\begin{align}\lim_{x\to 0^+}\ln(e^{0^+}-1)\\
\lim_{x\to 0^{+}} \ln(e^{0} - 1)\\
\lim_{x\to 0^{+}} \ln(1 - 1)\\
\lim_{x\to0^+}\ln0\end{align}
$\require{cancel}$
$\ln0=\cancel1$, right? But how is it $-\infty$?
Thank you
 A: No, $\ln(0)$ is undefined. First, consider that $e^x$ is the inverse function of $\ln(x),$ that is, $\ln(e^x)=x.$ The range of $e^x,$ is $(0,\infty).$ This indicates that $e^x\ne0,$ for any $x.$ Therefore $0$ is not in the domain of $\ln(x),$ so $\ln(0)$ is undefined.
As to the limit:
Since  $e^0=1,$ so $$\lim_{x^+\to 0} e^{x}-1=0.$$
We have that the composition of continuous functions is continuous (at least here we are continuous to the right of zero with $\ln$). So we get 
$$\lim_{x^+\to 0}\ln( e^{x}-1)=\lim_{y\to 0^+}~\ln(y)=-\infty.$$
Back to the domain and range stuff. The idea for this limit comes from
$$\lim_{x\to-\infty} e^x=0.$$
Which is true as a limit, but this doesn't mean $e^x$ is ever actually zero.
A: Just a side note as why $1$ isn't the right sided limit at $x=0$. 
In this case $lim_{x\to a+}=L $ then $ \forall \epsilon \gt 0 \  \exists \delta \ \gt 0 $ such that  $0 \lt x-a\lt \delta$ implies $|f(x)-L| \lt \epsilon$.
So if limit $L=1$ then we have $$0 \lt x \lt \delta \ \implies \ |ln(e^x-1)-1|\lt \epsilon$$
This means considered as sets, $\{x:0\lt x \lt \delta\}\  \subseteq \{x: |ln(e^x-1)-1|\lt \epsilon\}$
The second set can be expressed as $$\{ x: ln(1+e^{1-\epsilon}) \lt x \lt ln(1+e)\} \cup\{x: ln(1+x) \le x \lt ln(e^{\epsilon +1}+1) \}$$
But solving for $x$ we see $$\{x: 0 \lt x \lt \delta \} \nsubseteq \{ x: ln(e^{1-\epsilon}+1) \lt x \lt ln(e^{1+\epsilon}+1) \} \ \forall \epsilon \forall \delta$$
Thus $1$ cannot be the limit from the right at $x=0$.
