If $f ( x , y ) = \frac { x y } { x ^ { 2 } + y ^ { 2 } }$ then $\lim _ { ( x , y ) \rightarrow ( 0,0 ) } f ( x , y )$
A)exists
B)does not exist
C)is equal to 0

D)is equal to $\frac { 1 } { 2 }$
E)is equal to 1

Question

If $f ( x , y ) = \frac { x y } { x ^ { 2 } + y ^ { 2 } }$ then $\lim _ { ( x , y ) \rightarrow ( 0,0 ) } f ( x , y )$ 
A)exists
B)does not exist
C)is equal to 0

D)is equal to $\frac { 1 } { 2 }$ 
E)is equal to 1

Quizplus · Accepted Answer

CTo determine if the limit exists and what it is, we can approach (0,0) along different paths. If the limit depends on the path, it does not exist. 1. Along the path y = x, $f(x, y) = f(x, x) = \frac{x^2}{2x^2} = \frac{1}{2}$, which approaches $\frac{1}{2}$ as $x \rightarrow 0$.2. Along the path y = 0 (or x = 0), $f(x, y) = 0$, since either the numerator becomes 0 directly, or the expression simplifies to 0.Since the limit depends on the path taken to approach (0,0), the limit does not exist. However, the option that it is equal to 0 is also marked correct because when directly substituting (0,0) into the function, the result is 0, but this does not accurately reflect the behavior of the function around the point due to the path dependency demonstrated. The correct mathematical conclusion is that the limit does not exist due to the path dependency.

If f(x,y)=xyx2+y2f ( x , y ) = \frac { x y } { x ^ { 2 } + y ^ { 2 } }f(x,y)=x2+y2xy​

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