My problem is the following:
Let $\lambda>0$. Consider the function on $R^2$:
$f(x,y)=\frac{x^4+2y^2}{(x^2+y^2)^\lambda}$ for $(x,y)\ne (0,0)$ and $f(x,y)=0$ at $(x,y)=(0,0)$.
I now want to find the values of $\lambda>0$ such that $f(x,y)$ is continuous at $(0,0)$.
Correct me if I am wrong, but if I have understood it correctly, in order for $f(x,y)$ to be conti. at $(0,0)$, the following three must be true:
1, $f(0,0)$ is defined, (here given by $0$).
2, $\lim_{(x,y)\rightarrow (0,0)}f(x,y)$ exists.
3, $\lim_{(x,y)\rightarrow (0,0)}f(x,y)=f(0,0)(=0)$.
I first consider $\lambda\in ]0,1]$
I now choose a path $(t,m\cdot t)$, and see if the limit exists.
Calculations give that
$\lim_{(x,y)\rightarrow (0,0)}f(x,y)$=$\lim_{t\rightarrow 0}f(t,m\cdot t)$=$\frac{t^4+2m^2t^2}{(t^2+t^2)^\lambda}\leq \lim_{t\rightarrow 0}\frac{t^4+2m^2t^2}{(1+m^2)t^2}=\frac{2m^2}{1+m^2}$.
Therefore I concluded that there dosent exist a unique limit for the function $f$ in $\lambda \in ]0,1]$. Therefore I concluded that cannot be continuous at $(0,0)$ Is this correct concluded?
The second question I have, is to determine the $\lambda>0$ such that $f(x,y)$ becomes differentiable at $(0,0)$. I have no idea how to start on this, can anybody help me with that? If I find that there does not exist any $\lambda$ for which $f$ is continuous at $(0,0)$, is it then also fair to assume that there does not exist any $\lambda>0$ for which $f$ is differentiable at $(0,0)$?
Thanks in advance.
Best Answer
If $\lambda >1$ then $f(0,y)=\frac 2 {y^{2(\lambda -1)}}$ so $|f(0,y)| \to \infty $ as $y \to 0$. Hence $f$ is not continuous for $\lambda >1$.