f31d07…: “Unlike an autumn solution”

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Suppose that $S$ is a surface and the following is known.

The level curve corresponding to $z=1$ is $x^2+3y=4$ .
The level curve corresponding to $z=2$ is $2x^2+y=8$ .

From the given information, how many points $(a,b) \in \R ^2$ are there for which that $\lim _{\vector {x,y} \to \vector {a,b}} z(x,y)$ must not exist?

None One Two Three More than three, but finitely many Infinitely many

The points $(a,b) \in \R ^2$ for which $\lim _{\vector {x,y} \to \vector {a,b}} z(x,y)$ does not exist are

$\left (\answer {-2},\answer {0}\right ) \textrm { and } \left (\answer {2},\answer {0}\right )$ (in your response, list the point with the smaller $x$ -coordinate first)

The level curves are paths in the domain along which $z$ is constant. Since we have two different level curves, any point in the $xy$ -plane where they intersect will be a point on the surface for which there are multiple $z$ -values, and hence, where $\lim _{\vector {x,y} \to \vector {a,b}} z(x,y)$ will not exist.

For the level curve along which $z=1$ , we have $y = \answer {\frac {4}{3}-\frac {1}{3}x^2}$ .
For the level curve along which $z=2$ , we have $y = \answer {8-2x^2}$ .

Setting these equal gives $x=\answer {-2}$ and $x=\answer {2}$ (type the smaller $x$ -value first). Now, find the $y$ -values associated to each.