Question

Define h(0,0) in such a way that h is continuous at origin here n=5

h(u,v)=ln⁡((nu^2-u^2 v^2+nv^2)/(u^2+v^2 ))

Answer 1

Checking continuity at (0,0)

$\lim _{x\to \:0}(f(x))$

$\lim\limits_{u,v \to (0,0)}h(u,v)=\lim\limits_{u,v \to (0,0)} (\frac{ln⁡((nu^2-u^2 v^2+nv^2)}{(u^2+v^2 )})$

Left hand side

$\lim\limits_{u,v \to (0,0)} (\frac{ln⁡((5u^2-u^2 v^2+5v^2)}{(u^2+v^2 )})=-\infin$

Right-hand side

$\lim\limits_{u,v \to (0,0)} (\frac{ln⁡((5u^2-u^2 v^2+5v^2)}{(u^2+v^2 )})=-\infin$

h is not contiuous at (0,0)