If we define $f(x)= \sum_{m=0}^{\infty} x^{2^m} = \lim_{n\to\infty}f_n(x)= \lim_{n\to\infty}\sum_{m=0}^{n} x^{2^m}$.
It's easy to see that $|{f(x)}|$ converge in the interval $(-1,1)$. (We can easily born it with the geometric series)
Supose I want to look at the limit : $$\lim_{x\to1^-}|f(x)|$$ How could a prove that the limit diverge ?
I would like to evaluate the limit as x approach $1^-$ first and then use the limite as n tend to infinity like: $$ \lim_{x\to1^-}\lim_{n\to\infty}|f_n(x)| = \lim_{n\to\infty}\lim_{x\to1^-}|f_n(x)| $$
but my function is not continuous at x = 1. I also dont find any function that could born $f(x)$ ( that we know diverge at 1 and is smaller then $f(x)$ on the interval $(-1,1)$.