21ed25…: “Out of a gray bay”

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Consider the function $F(x,y)= \frac {2x^2+4xy^3+6y^6}{x^2+3y^6}$ .

As $\vector {x,y} \to \vector {0,0}$ along $x=0$ , $F(x,y)$ approaches $\answer {2}$ .
As $\vector {x,y} \to \vector {0,0}$ along $y=0$ , $F(x,y)$ approaches $\answer {2}$ .

Is this enough to determine that $\Lim {x,y}{0,0} = 0$ ? YesNo

As $\vector {x,y} \to \vector {0,0}$ along $y=x$ , $F(x,y)$ approaches $\answer {2}$ .

Is this enough to determine that $\Lim {x,y}{0,0} = 0$ ? YesNo

Maybe we’re a bit unlucky; let’s analyze along $y=mx$ .

As $\vector {x,y} \to \vector {0,0}$ along $y=mx$ , $F(x,y)$ approaches $\answer {2}$ .

Along $y=mx$ , we have

$F(x,y)= F\left (x,\answer {mx}\right ) = \frac {2x^2+4x\left (mx\right )^3+6\left (mx\right )^6}{x^2+3\left (mx\right )^6} = \frac {x^2 \cdot \left (\answer {2+4m^3x^2+6m^6x^4} \right ) }{x^2 \cdot \left (1 +3m^6x^4 \right ) }$

What’s happening here? By choosing a straight line path, the only coefficient that actually matters are those in front of the lowest power of $x$ . One thing we can notice is that by looking at a different type of path, we can try to balance the powers of $x$ and $y$ so each term has the same exponent. Let’s try $x=y^3$ .

As $\vector {x,y} \to \vector {0,0}$ along $x=y^3$ , $F(x,y)$ approaches $\answer {3}$ .

From this, we can conclude

$\Lim {x,y}{0,0} F(x,y)$ exists. $\Lim {x,y}{0,0} F(x,y)$ does not exist.

Take a minute to reflect on this; in order to determine that a limit exists, we have to show that the function tends to the same value along any path, not just check a few! Even if the function tends to the same value along every path of a certain type, you can never use the result to determine that a limit exists. Try to find part of a level curve as we did here; that is find a type of path that makes each term have the same power so cancellation occurs.