Determine if f is continuous at (0,0) where: f(x,y)={3x2yx4+2y2,(x,y)≠(0,0)0(x,y)=(0,0)

If a function f is continuous, then lim(x,y)→(a,b)f(x,y) is unique. Here, in this problem we will show that lim(x,y)→(0,0)f(x,y) is not unique. And hence, if is not continous. Solution: f(x,y)={3x2yx4+2y2,(x,y)≠(0,0)0(x,y)=(0,0) Let y→0 along the path y=mx2 where Now limx→0f(x,y)y=mx2 =limx→03x2yx4+2y2 =limx→03x2(mx2)x4+2(mx2)2 =limx→03mx4x4+2m2x4 =limx→03mx4x4x4x4+2m2x4x4 =limx→03m1+2m2 =3m1+2m2 Hence, for different values of m, the limit is different. Therefore lim(x,y)→(0,0)f(x,y) is not unique. Hence, f(x,y) is not continuous.

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Yasmin

Answered question

2021-10-31

Determine if f is continuous at (0,0) where:
$f (x, y) = {\begin{cases} \frac{3 x^{2} y}{x^{4} + 2 y^{2}}, & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases}$

Answer & Explanation

Anonym

Skilled2021-11-01Added 108 answers

If a function f is continuous, then $lim_{(x, y) \to (a, b)} f (x, y)$ is unique.
Here, in this problem we will show that $lim_{(x, y) \to (0, 0)} f (x, y)$ is not unique.
And hence, if is not continous.
Solution:
$f (x, y) = {\begin{cases} \frac{3 x^{2} y}{x^{4} + 2 y^{2}}, & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases}$
Let $y \to 0$ along the path $y = m x^{2}$ where
Now $lim_{x \to 0} f (x, y) y = m x^{2}$
$= lim_{x \to 0} \frac{3 x^{2} y}{x^{4} + 2 y^{2}}$
$= lim_{x \to 0} \frac{3 x^{2} (m x^{2})}{x^{4} + 2 {(m x^{2})}^{2}}$
$= lim_{x \to 0} \frac{3 m x^{4}}{x^{4} + 2 m^{2} x^{4}}$
$= lim_{x \to 0} \frac{3 m \frac{x^{4}}{x^{4}}}{\frac{x^{4}}{x^{4}} + 2 m^{2} \frac{x^{4}}{x^{4}}}$
$= lim_{x \to 0} \frac{3 m}{1 + 2 m^{2}}$
$= \frac{3 m}{1 + 2 m^{2}}$
Hence, for different values of m, the limit is different.
Therefore
$lim_{(x, y) \to (0, 0)} f (x, y)$ is not unique.
Hence, $f (x, y)$ is not continuous.

Jeffrey Jordon

Expert2022-06-23Added 2605 answers

Answer is given below (on video)

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