Answer :
Answer:
f(0,0)=ln19
Step-by-step explanation:
[tex] f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})[/tex] is given as continuous function, so there exist [tex]lim_{(x,y)\rightarrow(0,0)}f(x,y) [/tex] and it is equal to f(0,0).
Put x=rcosA annd y=rsinA
[tex]f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})[/tex]
we know that [tex]cos^2A+sin^2A=1[/tex], so we have that
[tex]f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)[/tex]
[tex]lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19[/tex]
So f(0,0)=ln19.