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

$Given, h(u,v)=ln⁡\frac{nu^2−u^2v^2+nv^2}{u^2+v^2​}.\newline \text{Solve for h,}\newline h(u,v)=ln⁡\dfrac{n(u^2+v^2)-(uv)^2}{u^2+v^2 }\newline =\ln⁡(\dfrac{n(u^2+v^2)}{u^2+v^2 }-\frac{(uv)^2}{u^2+v^2 })\newline =\ln⁡(n-\frac{(uv)^2}{u^2+v^2 })\newline \text{ Taking limit.}(u,v)\to(0,0)\newline lim_{(u,v)\to(0,0)} h(u,v)\newline =lim_{(u,v)\to(0,0)} \ln⁡(n-\frac{(uv)^2}{u^2+v^2 })\hspace{1cm}(\frac{0}{0} form)\newline (put \space v=mu.)\newline =lim_{u\to0} \ln⁡(n-\frac{u^4m^2}{u^2(1+m^2) })\newline =lim_{u\to0} \ln⁡(n-\frac{u^2}{(1+m^2) })\newline =ln(n)\newline \text{Thus,} h(x,y)=ln(n), at (x,y)=(0,0).$