Consider the function π(π₯, π¦) = π₯π¦ ( π₯ 2βπ¦ 2 π₯ 2+π¦2 ) , (π₯, π¦) β (0, 0). Show that lim (π₯,π¦)β(0,0) π(π₯, π¦) = 0
Convert to to polar coordinates: "x=r\\cos(\\theta), y=r\\sin (\\theta)"
"=\\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}"
"=\\dfrac{1}{2}(0)^2 \\sin(2\\theta)\\cos(2\\theta)=0"
Therefore
Comments
Leave a comment