Answer to Question #114434 in Calculus for Rajeev

Question

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

Answer 1

Consider

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

Suppose $\varepsilon>0$

$$\bigg|{6x^2y^3 \over x^2+y^3}\bigg|=6\bigg|{y^3 \over x^2+y^3}\bigg|x^2$$

Note that

$$\bigg|{y^3 \over x^2+y^3}\bigg|\leq1, x^2\leq x^2+y^2=\delta^2$$

So

$$6\bigg|{y^3 \over x^2+y^3}\bigg|x^2<6\cdot1\cdot\delta^2$$

If we choose $\delta=\sqrt{\dfrac{\varepsilon}{6}}$ then

$$6\bigg|{y^3 \over x^2+y^3}\bigg|x^2<6\cdot1\cdot\delta^2=\varepsilon$$

$$|f(x,y)-0|<\varepsilon, if\ \sqrt{x^2+y^2}<\delta$$

Hence

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