Answer to Question #105338 in Calculus for SHIVAM KUMAR

Question

check whether the limit of the function f (x,y)=3x^3y/x^6+2y ^2 exists as (x,y) to(0,0)

Answer 1

Let $f(x,y)=\dfrac{3x^3y}{x^6+2y^2}.$ Let's approach $(0,0)$ along the $x-$axis.

Then $y=0$ gives $f(x,0)=\dfrac{3x^3(0)}{x^6+2(0)^2}=0$ for all $x\not=0.$

$f(x,y)\to0$ as $(x,y)\to(0,0)$ along the $x-$axis.

Let's approach $(0,0)$ along the line $y=x^3.$

Then $y=x^3$ gives

$$f(x,x^3)=\dfrac{3x^3(x^3)}{x^6+2(x^3)^2}=1$$

$f(x,y)\to1$ as $(x,y)\to(0,0)$ along the line $y=x^3.$

Since $f$ has two different limits along two different lines, the given limit

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

does not exist.