f886c5…: “Towards a windless line”

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Consider the function $f(x,y) = \frac {2x^3y-4x^2y^2}{2x^4+y^4}$ .

The domain of this function is $\R$ $\left \{ z \in \R \big | z \neq 0 \right \}$ $\R ^2$ $\left \{ (x,y) \in \R ^2 \big | (x,y) \neq (0,0) \right \}$ .

We now examine whether $\Lim {\vector {x,y}}{\vector {0,0}} f(x,y)$ exists. By noting that the degree of each term in the numerator and denominator are the same, we can try to choose a type of path along which we can induce algebraic cancellation.

Which type of path would likely be a good choice?

lines planes parabolas circles

Let’s try to analyze along two specific lines.

Along $x=0$ , we find that
$f(x,y) = f(0,y) =\frac {2\left (\answer {0}\right )^3y-4\left (\answer {0}\right )^2y^2}{2\left (\answer {0}\right )^4+y^4} = \answer {0}, \textrm { for } y \neq 0.$
Thus, $f(x,y) \to \answer {0}$ as $\vector {x,y} \to \vector {0,0}$ along the path $x=0$ .
Along $y=0$ , we find that
$f(x,y) = f\left (\answer {x},\answer {0}\right ) = \answer {0} , \textrm { for } x \neq 0.$
Thus, $f(x,y) \to \answer {0}$ as $\vector {x,y} \to \vector {0,0}$ along the path $y=0$ .

Is this enough to conclude that $\Lim {\vector {x,y}}{\vector {0,0}}$ exists?

Yes No

Note that if $\Lim {\vector {x,y}}{\vector {0,0}} f(x,y)$ exists, the above shows that it must be $0$ ; that is, along every path, the function must tend to $0$ .

Let’s pick another explicit path, say $y=2x$ .

$f(x,y) = f\left (x,\answer {2x}\right ) =\frac {2x^3\left (\answer {2x}\right )-4x^2\left (\answer {2x}\right )^2}{2x^4+\left (\answer {2x}\right )^4} = \frac {\answer {-12}\cdot x^4}{\answer {18}\cdot x^4} = -\frac {2}{3}, x \neq 0.$

Thus, $f(x,y) \to \answer {-\frac {2}{3}}$ as $\vector {x,y} \to \vector {0,0}$ along the path $y=2x$ .

Is this enough to conclude that $\Lim {\vector {x,y}}{\vector {0,0}}$ exists or does not exist?

Yes; $\Lim {\vector {x,y}}{\vector {0,0}}$ exists. Yes; $\Lim {\vector {x,y}}{\vector {0,0}}$ does not exist. No

Since we think that analyzing the outputs of the function along lines in its domain, we can require that $y=mx$ and see what happens.

Along $y=mx$ , we have the following.

$f(x,y) = f\left (x,\answer {mx}\right ) =\frac {2x^3\left (\answer {mx}\right )-4x^2\left (\answer {mx}\right )^2}{2x^4+\left (\answer {mx}\right )^4} = \frac {\answer {2m-4m^2}\cdot x^4}{\answer {2+m^4}\cdot x^4} = \frac {\answer {2m-4m^2}}{2+m^4}, x \neq 0.$

Thus, $f(x,y) \to \frac {\answer {2m-4m^2}}{2+m^4}$ as $\vector {x,y} \to \vector {0,0}$ along the path $y=mx$ .

Is this enough to conclude that $\Lim {\vector {x,y}}{\vector {0,0}}$ exists or does not exist?

Yes; $\Lim {\vector {x,y}}{\vector {0,0}}$ exists because $f(x,y)$ tends to a number along each path. Yes; $\Lim {\vector {x,y}}{\vector {0,0}}$ does not exist because $f(x,y)$ approaches a different value for different choices of $m$ . No

The results of both approaches are related.

The path $y=0$ is obtained from $y=mx$ by setting $m = \answer {0}$ .
The path $x=0$ is is a vertical line, so we take $m \to \infty$ .
The path $y=2x$ is obtained from $y=mx$ by setting $m = \answer {2}$ .
Do the results of both agree?
One nice consequence of analyzing the function along paths of the form $y=mx$ is that this gives explicit information about how the outputs of the function depend on the choice of path.