Assume that $\lim\_{(nt) |z|\to 1}|f(z)|(1-|z|)^p=0$, where $f$ is analytic in the unit disk and $p>0$,where $(nt)|z|\to 1$ nontangentially. Does this implies that $\lim\_{|z|\to 1}|f(z)|(1-|z|)^p=0$ uniformly?



Holder class of analytic functions
No, it does not. There can be a region $D$ in the disc with one boundary point on the circle,
say $1$, which approaches this point tangentially. And a function which is bounded outside
$D$ but grows arbitrarily fast in $D$. To non-tangential limits of your expression at all points are zero, but there is no uniformity, and the growth is arbitrarily high.
Such function can be constructed in many ways, using for example approximation theorems,
or Cauchy integrals.


