9caae6…: “Throughout the winter eyes”

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Consider $\lim _{\vector {x,y}\to \vector {0,0}}\frac {2x^2+3 x y}{4y^2-3xy}$ .

To determine if this limit exists, consider the outputs of the function $F(x,y) = \frac {2x^2+3 x y}{4y^2-3xy}$ along the line $y = m x$ .

When $m = 1$ , $F(x,y)$ approaches $\answer {5}$ as $\vector {x,y} \to \vector {0,0}$ along $y= x$ .
When $m = 2$ , $F(x,y)$ approaches $\answer {4/5}$ as $\vector {x,y} \to \vector {0,0}$ along $y= 2 x$ .
Leaving $m$ unspecified, $F(x,y)$ approaches $\answer {\frac {2+3m}{4m^2-3m}}$ as $\vector {x,y} \to \vector {0,0}$ along $y= m x$ .

Does the limit exist? If it does, indicate its value and if it does not exist, write $\verb |DNE|$ .

$\lim _{\vector {x,y}\to \vector {0,0}}\frac {2x^2+3 x y}{4y^2-3xy} = \answer {DNE}$

Note that if the limit exists, the function must approach the same value along any path as $\vector {x,y} \to \vector {0,0}$ . we can conclude that the limit does not exist by noting either of the following.

The function approaches a different value as $\vector {x,y} \to \vector {0,0}$ along the path $y=x$ than it does as $\vector {x,y} \to \vector {0,0}$ along the path $y=2x$ .
The value the function approaches as $\vector {x,y} \to \vector {0,0}$ along the path $y=mx$ depends on the choice of $m$ (and thus on the choice of straight line path).