Limit of function composition $\lim_{x \to 0}\ln(1+x)^{\frac{1}{x}} = \ln [\lim_{x \to 0}(1+x)^{\frac{1}{x}}]$ $$\lim_{x \to 0}\ln(1+x)^{\frac{1}{x}} = \ln [\lim_{x \to 0}(1+x)^{\frac{1}{x}}]$$
While attempting to understand the approximation of $e$, I ended up at the above step where the RHS is derived from the following theorem. 
If $f$ is continuous at $c$ and  $\lim_{x\rightarrow b} g=c$,  we have $\lim_{x\rightarrow b} f\circ g= f( \lim_{x\rightarrow b} g)$. 
Here, $f= \ln x$,  $g=(1+x)^{\frac{1}{x}}$ and $b=0$. As $x\rightarrow b$, $g\rightarrow 1^\infty$. I have read online that one raised to the power of Infinity( and not undefined as $x\neq b$ ) is indeterminate. If that is true, $\ln$ cannot be continuous at $c$( indeterminate ). How can we then take the limit as shown in the picture?

Does, "If $f$ is continuous at $c$ and  $\lim_{x\rightarrow b} g=c$,  we have $\lim_{x\rightarrow b} f\circ g= f( \lim_{x\rightarrow b} g)$" hold if $f$ is not continuous at $c$ but its limit is c?
I believe it does as per the formal definition of the limit. 
 A: This is a very common confusion. Here's the thing: "indeterminate" does NOT mean "undefined" (or "does not exist"). Roughly speaking, we say that a limit is an indeterminate form if from the parts of the expression we can't automatically deduce the value of the limit. For example:


*

*$\frac{6}{3}$ is not an indeterminate form, because it's definitely equal to $2$. This means that if you have a limit $\lim\limits_{x\to a}\frac{f(x)}{g(x)}$, where $\lim\limits_{x\to a}f(x)=6$ and $\lim\limits_{x\to a}g(x)=3$, then you can for sure conclude that $\lim\limits_{x\to a}\frac{f(x)}{g(x)}=2$.

*But $\frac{0}{0}$ is an indeterminate form. This means that if you have a limit $\lim\limits_{x\to a}\frac{f(x)}{g(x)}$, where $\lim\limits_{x\to a}f(x)=0$ and $\lim\limits_{x\to a}g(x)=0$, then the result can be different depending on the specific functions $f$ and $g$. Note that I did NOT say "undefined"! Such a limit may turn out to be undefined, or to be equal to any real number, or to be infinite. For example:
$$\lim\limits_{x\to0}\frac{x^2}{x^4}=+\infty, \quad \lim\limits_{x\to0}\frac{x^2}{x^2}=1, \quad \lim\limits_{x\to0}\frac{x^4}{x^2}=0, \quad \lim\limits_{x\to0}\frac{18x^2}{2x^2}=9, \quad \text{etc,}$$
where all these limits, if you first look at their numerators and denominators separately, are of the form $\frac{0}{0}$.
Another way to rephrase this: calling a limit "indeterminate" is NOT the answer! It only means that we can't deduce the answer by (roughly speaking) basic observation. At this point the answer is still unknown to us; so we have to do some more work, possibly using more advanced techniques, to find the limit (or demonstrate that it doesn't exist).
As you correctly pointed out, $1^{\infty}$ is an indeterminate form. But you misinterpreted what that means. It only means that we don't know yet the answer to this particular limit (as opposed to seeing something like $2^3$, when we'd immediately conclude that the answer is $8$). We still have no idea whether $c$ exists or not! For this reason, we don't know yet whether the theorem applies here or not.
So we have to do some more work to find the inside limit $\lim\limits_{x\to0}(1+x)^{1/x}$. One way or another — (insert some solution here) — we can find that actually $\lim\limits_{x\to0}(1+x)^{1/x}=e$. So our $c=e$, and since $f(x)=\ln x$ is continuous at $c=e$, now we know that the theorem is perfectly applicable here.
A: Note that
$$\ln[(1+x)^{1/x}]=\frac{\ln(1+x)}x.$$
One really shouldn't apply L'Hospital's rule to this, but what happens if one does?
