Answer to Question #115219 in Calculus for Jeetu jat

Question

Check whether the limit of the function 6 2
3
2
3
( , )
x y
x y
f x y
+
= exists as )0

Answer 1

$$\lim\limits_{(x,y)\to (0,0)}{6x^2y^3\over x^2+y^3}$$

Let $x=r\cos\theta, y=r\sin\theta$

$$\lim\limits_{(x,y)\to (0,0)}{6x^2y^3\over x^2+y^3}=\lim\limits_{r\to 0}{6r^2\cos^2\theta r^3\sin^3 \theta\over r^2\cos^2\theta+ r^3\sin^3 \theta }=$$

$$=\lim\limits_{r\to 0}{6r^3\cos^2\theta \sin^3 \theta\over \cos^2\theta+ r\sin^3\theta }={6(0)^3\cos^2\theta \sin^3 \theta\over \cos^2\theta+ (0)\sin^3\theta }=0$$

$$\lim\limits_{(x,y)\to (0,0)}f(x,y)=\lim\limits_{(x,y)\to (0,0)}{6x^2y^3\over x^2+y^3}=0$$