Show that $\lim_{z\to 0}|z|^{\sqrt 2}f(z)=0$ in $D$ 
Let , $f$ be analytic in $\mathbb D\setminus\{0\}$ and unbounded near $z=0$. If the function $|z|^{\sqrt 2}f(z)$ is bounded at $z=0$ then show that $$\lim_{z\to 0}|z|^{\sqrt 2}f(z)=0\text{  &  }\lim_{z\to 0}|z|^{\sqrt 2/2}f(z)=\infty.$$where , $\mathbb D=\{z\in \mathbb C:|z|<1\}.$

Since, the function $|z|^{\sqrt 2}f(z)$ is bounded at $z=0$ so, it has a removable singularity at $z=0$ and so, $\lim_{z\to 0}z.|z|^{\sqrt 2}f(z)=0$. From here , how I can show the required limit ?

Hint: $z^2f(z)$ is analytic in some $\{0<|z|<r\},$ and from the given information, $z^2f(z) \to 0$ at $0.$ In particular $z^2f(z)$ is bounded in this domain. Hence it has a removable singularity at $0.$ Thus $z^2f(z) = \sum a_nz^n $ in this domain. Think about the first few of the coefficients $a_n.$
