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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

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    $\begingroup$ $\ln1=0$ and $\ln0=-\infty$ $\endgroup$ May 10 '17 at 14:00
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    $\begingroup$ Recall that $\ln 0$ is undefined. It is not 1. $\endgroup$ May 10 '17 at 14:00
  • $\begingroup$ I believe you confused logarithm and exponential. $\endgroup$ May 10 '17 at 14:00
  • $\begingroup$ @ParclyTaxel thank you. So my computation was right? except for the last part? Also, ln(0) = -∞ but is this only when we are taking limits? Say we are just given the function f(0) = ln(x) that would be undefined? but since we are in a limit it's going to be -∞ solely because it's inside a limit? $\endgroup$
    – yre
    May 10 '17 at 14:05
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ May 10 '17 at 14:06
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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.

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    $\begingroup$ Thank you. Isn't the function ln(0) undefined, but when it's inside a limit either approaching left side or right side or infinity then it's -∞? So if it were a say f(0) = ln(x) that's UND, but if it's lim x->∞ ln(0) then it's -∞ ? $\endgroup$
    – yre
    May 10 '17 at 14:10
  • $\begingroup$ $\ln(0)$ is undefined, but this doesn't meant that the limit is undefined. A limit represents the quantity which your function is approaching as $x\to 0^+,$ not the value it ever actually takes on. $\endgroup$ May 10 '17 at 14:11
  • $\begingroup$ Is it because e^x is really something like e^(0.01), e^(0.001), e^(0.001), e^(0.0001),.... so it's getting very close to 0, therefore e^x is getting super to close to 1, but not quite 1. Therefore (super close to 1 MINUS 1) means we are consistently approach something very small, or therefore approaching 0, and this is the limit process, it's approaching 0 and went ln(x) approaches 0 that means it's approaching -∞? $\endgroup$
    – yre
    May 10 '17 at 14:17
  • $\begingroup$ Yes, that's the idea of a limit. $\endgroup$ May 10 '17 at 14:18
  • $\begingroup$ So I can see it visually like: 0.01 - 1, 0.001 - 1, 0.0001 - 1, 0.00001 - 1,.... So each increment closer, we are in turn getting close to 0, as the limit gets consistently closer to 0 the graph of ln(x) is in turn getting consistently closer to -∞? $\endgroup$
    – yre
    May 10 '17 at 14:19
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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 $x=0$.

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