\magnification=\magstep1
\centerline{Mathematics 41, section 1 -- Multivariable Calculus}
\centerline{Discussion 3 -- Limits for $f(x,y)$}
\centerline{October 13, 1999}
\bigskip
\noindent
{\bf Background}
\bigskip
As is implied by their name, the {\it partial derivatives\/} of 
a function $f(x,y)$ are not the ``whole story'' to understanding
how $f$ is changing at $(x_0,y_0)$ in its domain.  We also want to 
develop something like a ``total derivative'' of $f$ at $(x_0,y_0)$,
and to do this we need to understand {\it limits\/} of functions
$f(x,y)$ as $(x,y) \to (x_0,y_0)$.  In other words, we want to 
understand what it should mean to say 
$$\lim_{(x,y) \to (x_0,y_0)} f(x,y) = L$$
for some real number $L$.  
\bigskip
\noindent
{\bf Discussion Questions}
\bigskip
A first observation is that we can let 
$(x,y) \to (x_0,y_0)$ in many ways in ${\bf R}^2$.  For instance,
given a direction vector $\vec{v} = (v_1,v_2)$, we could
let $(x,y) = (x_0 + t v_1, y_0 + t v_2)$ (on the line through
$(x_0,y_0)$ in the direction of $\vec{v}$), and take an ordinary
limit as $t\to 0$:
$$\lim_{t\to 0} f(x_0 + t v_1, y_0 + t v_2)\leqno(1)$$ 
\bigskip
\item{A)}  For instance, what happens if you let $f(x,y) = x^2 - 2xy + 4y^2$,
$(x_0,y_0) = (1,2)$, and take a limit as in (1) for several different
$\vec{v}$?  Does the choice of $\vec{v}$ seem to matter?  Explain.
Does it make to sense to say $\lim_{(x,y) \to (1,2)} f(x,y) = L$
for some $L$?  If so, which $L$ ``works''?  If not, why not?
\bigskip
\item{B)}  Now, suppose 
$$g(x,y) = \cases{ {xy\over x^2 + y^2} & if $(x,y) \ne (0,0)$\cr
                          0 & if $(x,y) = (0,0)$\cr}$$
\itemitem{1)} What happens if you proceed as in A for $(x_0,y_0) = (1,1)$
and take several different $\vec{v}$?  Does the choice of $\vec{v}$ 
seem to matter?  Explain.
\itemitem{2)} Does it make to sense to say $\lim_{(x,y) \to (1,1)} g(x,y) = L$
for some $L$?  If so, which $L$ ``works''?  If not, why not?
\itemitem{3)} Now repeat part 1 for $(x_0,y_0) = (0,0)$.  Is 
this case different?  Why or why not.  You may want to refer to the
graph and contour plot of $g(x,y)$ on the accompanying Maple
worksheet.
\itemitem{4)} Does it make to sense to say $\lim_{(x,y) \to (0,0)} g(x,y) = L$
for some $L$?  If so, which $L$ ``works''?  If not, why not?  
\bigskip
\item{C)}  Now, let 
$$h(x,y) = \cases{ {y^2 - x^4 \over y^2 + x^4} & if $(x,y) \ne (0,0)$\cr
                                0 & if $(x,y) = (0,0)$\cr}$$
\itemitem{1)} What happens if you proceed as in A for $(x_0,y_0) = (0,0)$
and take several different $\vec{v}$?  Does the choice of $\vec{v}$ 
seem to matter?  Explain.  Refer to the graph and contour
plot of $h(x,y)$ on the accompanying Maple worksheet.
\itemitem{2)}  Before you decide whether $\lim_{(x,y) \to (0,0)} h(x,y)$
should exist, also consider letting $(x,y) \to (0,0)$ by letting
$t\to 0$ on 
the following two parabolas: $(x,y) = (t,t^2)$ and $(x,y) = (t,2t^2)$.
and computing 
$$\lim_{t\to 0} h(t,t^2)$$
and 
$$\lim_{t\to 0} h(t,2t^2).$$
\itemitem{3)} Does it make to sense to say $\lim_{(x,y) \to (0,0)} h(x,y) = L$
for some $L$?  If so, which $L$ ``works''?  If not, why not?   
\bigskip
\item{D)}  What should it mean to say $\lim_{(x,y)\to (x_0,y_0)} f(x,y) = L$?
\bigskip
\noindent
{\bf Assignment}
\bigskip
Group write-ups due Monday, October 18.
\bye
