Determine the set of points at which the function is continuous. f(x,y)=x2y32x2+y2 if (x,y) not equal to (0,0) and 1 if (x,y)=(0,0)

On R2 without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point except the origin, we must check whether lim(x,y)→(0,0)f(x,y)=f(0,0) Checking along the line y=0, we see that limx→0x2(0)32x2+(0)2=limx→00=0 But this implies that if f has a limit as (x,y) tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this cse, f(x,y) is not continuous at the origin.

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Algebra
Pre-Algebra
Negative numbers and coordinate plane

tearstreakdl

Answered question

2021-12-13

Determine the set of points at which the function is continuous.

f (x, y) = x^{2} \frac{y^{3}}{2} x^{2} + y^{2}

if (x,y) not equal to (0,0) and 1 if (x,y)=(0,0)

Answer & Explanation

Jordan Mitchell

Beginner2021-12-14Added 31 answers

R^{2}

without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point except the origin, we must check whether

lim_{(x, y) \to (0, 0)} f (x, y) = f (0, 0)

Checking along the line

y = 0

, we see that

lim_{x \to 0} \frac{x^{2} {(0)}^{3}}{2 x^{2} + {(0)}^{2}} = lim_{x \to 0} 0 = 0

But this implies that if f has a limit as

(x, y)

tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this cse, f(x,y) is not continuous at the origin.

Edward Patten

Beginner2021-12-15Added 38 answers

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