Why can they use continuity to compute this limit? 

(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?
 A: 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, $\lim_{x\to 0} \ln((1+x)^{1/x})=1$
A: 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.
