Trying the solve lim x->0 (1+1/x)^x with L'Hôpital's rule

[pineapplewithmouse]

I want to solve the limit lim x->0 (1+x)^1/x from scratch (so assuming I don't know it's equal to e).
This limit has the indeterminate form of 1^infinity, and according to Wikipedia, you can transform 1^infinity into e to the power of 0/0 (or infinity/infinity for that matter), and then you can use L'Hôpital's rule on the limit that has the 0/0 form.
As I understand it, the transformation was done in this way:

And then:

So the equality in the second picture is a particular case of the equality:

And as I understand it, this equality hold true if and only if p(x) is continuous, and if the limit lim x -> c q(x) exists (i.e. is a real number).

But when I solve the limit from scratch, I don't actually know the answer, so I don't know if the limit lim x -> c q(x) exists or not (in this case, the limit lim x -> 0 (ln(1+1/x))/(1/x)), so how can I make the transformation above? Is there some indication that the limit does exist? Or am I misunderstanding something about the transformation?

 

[fresh_42]

Let us assume that [imath] \displaystyle{ \lim_{x \to c} q(x) } [/imath] does not exist. Then how can [imath] p(x) [/imath] be continuous at [imath] q(c) [/imath]?

 

[Dr.Peterson]

But when I solve the limit from scratch, I don't actually know the answer, so I don't know if the limit lim x -> c q(x) exists or not (in this case, the limit lim x -> 0 (ln(1+1/x))/(1/x)), so how can I make the transformation above?

You try to find the limit of q(x). If it exists, you can use it. If it doesn't exist, then the limit you want doesn't exist.

What have you done to find it?