Answer to Question #186264 in Calculus for noman

Question

Define h(0,0) in such a way that h is continuous at origin h(u,v)=ln⁡((nu^2-u^2 v^2+nv^2)/(u^2+v^2 ))

Answer 1

$h(u,v)=\ln⁡\dfrac{nu^2-u^2 v^2+nv^2}{u^2+v^2 }$

$h(u,v)=\ln⁡\dfrac{nu^2+nv^2-u^2 v^2}{u^2+v^2 }$

$h(u,v)=\ln⁡\dfrac{n(u^2+v^2)-(uv)^2}{u^2+v^2 }$

$h(u,v)=\ln⁡(\dfrac{n(u^2+v^2)}{u^2+v^2 }-\dfrac{(uv)^2}{u^2+v^2 })$

$h(u,v)=\ln(⁡{n}- \dfrac{(uv)^2}{u^2+v^2 })$

$\lim\limits_{u,v \to 0,0}= \ln n$

$\therefore$ h(u,v) is not continuous at origin.