Prove that, if $w$ is a $(2^N)$th root of unity, where $N \in \mathbb N$, then:
$$\lim_{r\to 1^-}|f'(rw)| = \infty$$
Where:
$$f(z) = \sum\limits_{j = 1}^\infty 2^{-j}z^{2^j}$$
I haven't done left and right limits in so long, that I just don't know where to begin.
Clearly $f$ is analytic in the unit disc and $f'(z)=\sum_{j=1}^\infty z^{2^j}$, where $|z|<1$.
Let $s_n(z)=\sum_{j=1}^n z^{2^j}$. Then for $z=rw=r\exp(2k\pi i/2^{-N})$, we have that $w^{2^j}=1$, for $j\ge N$, and thus $$ s_n(rw)=\sum_{j=1}^{N-1} (rw)^{2^j}+\sum_{j=N}^n (rw)^{2^j}=\sum_{j=1}^{N-1}(rw)^{2^j}+\sum_{j=N}^{n} r^j=\sum_{j=1}^{N-1}(rw)^{2^j}+\frac{r^{N}-r^{n+1}}{1-r} \to \sum_{j=1}^{N-1}(rw)^{2^j}+\frac{r^{N}}{1-r} $$ as $n\to \infty$. Thus $$ f'(rw)=\sum_{j=1}^{N-1}(rw)^{2^j}+\frac{r^{N}}{1-r}, $$ and hence $$ \lim_{r\to 1^-}\lvert f'(rw)\rvert=\infty. $$