Answer to Question #226881 in Calculus for BINGIWE MAHLANGU

Question #226881

Consider the function 𝑓(π‘₯, 𝑦) = π‘₯𝑦 ( π‘₯ 2βˆ’π‘¦ 2 π‘₯ 2+𝑦2 ) , (π‘₯, 𝑦) β‰  (0, 0). Show that lim (π‘₯,𝑦)β†’(0,0) 𝑓(π‘₯, 𝑦) = 0


1
Expert's answer
2021-08-17T18:16:06-0400

Convert to to polar coordinates: "x=r\\cos(\\theta), y=r\\sin (\\theta)"


"\\lim\\limits_{(x,y)\\to(0,0)}\\dfrac{xy(x^2-y^2)}{x^2+y^2}"

"=\\lim\\limits_{r\\to0}\\dfrac{r\\cos(\\theta) r\\sin(\\theta)(r^2\\cos^2(\\theta)-r^2\\sin^2(\\theta))}{r^2\\cos^2(\\theta)+r^2\\sin^2(\\theta)}"

"=\\lim\\limits_{r\\to0}\\dfrac{r^4\\cos(\\theta) \\sin(\\theta)(\\cos^2(\\theta)-\\sin^2(\\theta))}{r^2}"


"=\\lim\\limits_{r\\to0}\\dfrac{1}{2}r^2 \\sin(2\\theta)\\cos(2\\theta)"

"=\\dfrac{1}{2}(0)^2 \\sin(2\\theta)\\cos(2\\theta)=0"

Therefore


"\\lim\\limits_{(x,y)\\to(0,0)}\\dfrac{xy(x^2-y^2)}{x^2+y^2}=0"



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