A Question about solution of Laplace's equation in p.d.e Let $u:\mathbb{R}^2\setminus\{(0,0)\}\rightarrow \mathbb{R}$ be a $C^2$  function satisfying $\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=0$,for all $(x,y)\neq (0,0)$ suppose  $u$ is the form  $u(x,y)=f(\sqrt{x^2+y^2})$, where $f:(0,\infty)\rightarrow \mathbb{R}$ is non constant function , then
$1. \lim _{x^2+y^2\rightarrow 0}|u(x,y)|=\infty$
$2. \lim _{x^2+y^2\rightarrow 0}|u(x,y)|=0$
$3. \lim _{x^2+y^2\rightarrow \infty}|u(x,y)|=\infty$
$4. \lim _{x^2+y^2\rightarrow \infty}|u(x,y)|=0$
My attempt: 
solution of the Laplace's equation with no boundary conditions:
so general solution of the this laplace's equation 
but how to solve this question 

$u(x,y)=f(\sqrt{x^2+y^2})\\
u_x = \frac{x}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2})\\u_{xx} = \frac{\left[\sqrt{x^2+y^2}\left[\frac{x^2}{\sqrt{x^2+y^2}}f''(\sqrt{x^2+y^2})+f'(\sqrt{x^2+y^2}) \right]-\frac{x^2}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2})\right]}{(\sqrt{x^2+y^2})^2}\\u_{xx} = \frac{x^2 f''(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^2}+\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}-\frac{x^2 f'(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^3} \\ Let \; r=\sqrt{x^2+y^2}\;\; then \\u_{xx}=\frac{x^2f''(r)}{r^2}+\frac{f'(r)}{r}-\frac{x^2f'(r)}{r^3}\\ similarly \;\; u_{yy}=\frac{y^2f''(r)}{r^2}+\frac{f'(r)}{r}-\frac{y^2f'(r)}{r^3}\\$ 
hence the given Laplace equation $u_{xx}+u_{yy}=0$ becomes
$\frac{(x^2+y^2)f''(r)}{r^2}+\frac{2f'(r)}{r}-\frac{(x^2+y^2)f'(r)}{r^3}=0\\ \frac{r^2f''(r)}{r^2}+\frac{2f'(r)}{r}-\frac{r^2f'(r)}{r^3}=0\\ f''(r)+\frac{2f'(r)}{r}-\frac{f'(r)}{r}\\ f''(r)+\frac{f'(r)}{r}=0\\ \frac{f''(r)}{f'(r)}=-\frac{1}{r}$ 
on integration we get
$ \;\;log(f'(r))=-logr+loga \\ f'(r)=\frac{a}{r}$ 
on integrating again we get $\;\; f(r)=a\ logr+b $
f(r) is nonconstant if $a\ne 0 \\ as \;\;x^2+y^2\to 0, \;\;r^2\to 0 \Rightarrow r\to 0 \\ hence \;\; lim_{x^2+y^2\to 0} |u(x,y)|=lim_{r\to 0} |f(r)|=lim_{r\to 0}( a\ logr+b)=\infty\\ also \;\;lim_{x^2+y^2\to \infty} |u(x,y)|=lim_{r\to \infty} |f(r)|=lim_{r\to \infty}( a\ logr+b)=\infty $
option 1 and 3 are correct
