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Alexxandr [17]
3 years ago
5

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

0,0) be defined to be .
Mathematics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

f(0,0)=ln19

Step-by-step explanation:

f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2}) is given as continuous function, so there exist lim_{(x,y)\rightarrow(0,0)}f(x,y) and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

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)})

we know that cos^2A+sin^2A=1, so we have that

f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)

lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19

So f(0,0)=ln19.

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