You just have to be careful about one thing over here. You assumed the limit exists. How do you know the limit exists? Note, this is why Krizalid wrote "assuming it exists". To prove this we can note the function 0,\infty)\mapsto \mathbb{R}" alt="f0,\infty)\mapsto \mathbb{R}" /> defined as is increasing since . Therefore, the sequence is an increasing sequence. Now show that is bounded. So we have a increasing bounded sequence and therefore we have a limit.
That is not how I like to define . I like to define . And we can define to be the (unique) number such that . If we define this way then it would follow that . But whatever, it depends on your style of defining logarithmic functions. I just find that the approach I use it the cleanest and smoothest.
I will present this proof for what it's worth. Though, the posters proof builds
from this. This proof builds on the differentiability of ln(x). To be more exact
the
derivative of ln(x) at x=1.
Using the definition of a derivative, since 1/x=1, we get:
Therefore, it follows that:
from the continuity of can be written this way:
Now, we recognize this limit. The others play off it.
To show
merely let .
This changes the limit to and we have said limit.