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\leftline{Mathematics 241 -- Multivariable Calculus}
\leftline{Quiz 3 Solutions -- September 21, 2007}
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\noindent
A) (5)  Sketch the level curves with $c = -1,0,1$ for the function
$f(x,y) = x^2 - y^2$.  
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{\it Solution:}  The level curve for the value $c$ is $x^2 - y^2 = c$.
When $c = 0$, we get the two lines $y = \pm x$.  With $c \ne 0$, the 
level curve is a hyperbola.  The hyperbola opens left and right if $c > 0$.
The hyperbola opens up and down if $c < 0$.  Here is a plot, from Maple:
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\noindent
B) (5)  Let 
\[
f(x,y) = \begin{cases} 
\frac{x + y}{\sqrt{x^2 + y^2}} & \text{if} (x,y) \ne (0,0)\\
                    0 & \text{if} (x,y) = (0,0)
\end{cases}
\]
Does $f$ have a limit as $(x,y) \to (0,0)$ along the line $y = 0$?  
(Be careful, $+\sqrt{a^2} = |a|$, not
$a$.) What does your answer say about the existence
of $\lim_{(x,y) \to (0,0)} f(x,y)$?
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{\it Solution}.  If $y = 0$, then $f(x,0) = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|}$, 
which is $+1$ if $x > 0$ and $-1$ if $x < 0$.  This says,
\[
\lim_{x \to 0^-} f(x,0) = -1 \ne 1 = \lim_{x \to 0^+} f(x,0).
\]
Since there is not even a single limit along this one line, the limit
of $f(x,y)$ as $(x,y) \to (0,0)$ cannot exist.  
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