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(a) Use l'Hôpital's rule to find $\displaystyle\lim_{x\to0}\ln[(1+x)^{1/x}]$.
Solution: Since $\ln x$ is continuous, $$\lim_{x\to0}\ln\left((1+x)^{1/x}\right)=\ln\left(\lim_{x\to0}(1+x)^{1/x}\right)=1.$$ Therefore, $\displaystyle\lim_{x\to0}(1+x)^{1/x}=e$.

In the solution above, they've reasoned that because "$\ln(x)$" is continuous, they are able to use limit laws to directly take the limit of the inner function. However, $\ln x$ is undefined and has a vertical asymptote at $x=0$, so how are they able to still use continuity (and thus direct substitution) to compute this limit? If I were not allowed to use graphing software and had no idea how the graph of $\ln\left[(1+x)^{1/x}\right]$ looked like, how would I know that I can use continuity?

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I think what is confusing you is the last line, which seems wrong to me, or to be more specific at a wrong place. Also this is not an application of L'Hospital's rule.

Yes, $\ln$ is not defined for $x=0$, but this value never accours in this calculation. The continuouity is used to pull the limit in. Then it is evaluated, that

$\lim_{x\to 0} (1+x)^{1/x}=e$. (Do you see how?)

And now $\ln(e)=1$.

The last line should read like this,

Therefore, $\lim_{x\to 0} \ln((1+x)^{1/x})=1$

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  • $\begingroup$ I think you're correct, but I think "Do you see how?" is hiding some circular reasoning. Or is there some other way to show that limit? Or are you taking that limit to be your definition of e? $\endgroup$
    – Teepeemm
    Jan 1, 2023 at 20:42
  • $\begingroup$ @Teepeemm I use $\lim_{n\to\infty} (1+1/n)^n$, as definition for $e$. My question refered to the substituton of $1/n$ for $x$ and let $x\to 0$ instead of $n\to\infty$. $\endgroup$
    – Cornman
    Jan 2, 2023 at 9:12
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Here's my interpretation of their solution:

We know that $\displaystyle\lim_{x\to0}(1+x)^{1/x}=e$. Since $\ln x$ is continuous at e, $$\lim_{x\to0}\ln((1+x)^{1/x})=\ln(\lim_{x\to0}(1+x)^{1/x})=\ln(e)=1.$$ As a bonus, because $e^x$ is continuous at 1, this shows that $\displaystyle\lim_{x\to0}(1+x)^{1/x}=e^1=e$.

To answer your original question: you don't need ln to be continuous everywhere, just at the argument to ln.

My complaint about this approach is a little more apparent with this write up than the original: how do we know that $\displaystyle\lim_{x\to0}(1+x)^{1/x}=e$?

The usual way to show this is the following:

By properties of logarithms and then l'Hôpital's rule: $$\lim_{x\to0}\ln((1+x)^{1/x})=\lim_{x\to0}\frac{1}{x}\ln(1+x)=\lim_{x\to0}\frac{\ln(1+x)}{x}=\lim_{x\to0}\frac{\frac{1}{1+x}}{1}=1.$$ As a bonus, because $e^x$ is continuous at 1, this shows that $\displaystyle\lim_{x\to0}(1+x)^{1/x}=e^1=e$.

In fact, the usual reason that you're finding this limit is because you're interested in the bonus reason, and you use the logarithmic limit to derive the e limit.

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