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\centerline
{
\bf Dr. Z's Math251 Handout \#14.2 (2nd ed.)
[Limits and Continuity in Several Variables]
}

By Doron Zeilberger

{\bf Problem Type 14.2a}: Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (a,b)} f(x,y) \quad .
$$

{\bf Example Problem 14.2a}: 
Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (0,0)} {{2x^2} \over {3x^2+4y^2}}
\quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} If the function is {\it continuous} at the
point, which is the case if both top and bottom are
`nice' functions like polynomials or exponentials or
sines or cosines {\bf and} the denominator does not
vanish at the designated point $(a,b)$, then just
plug-it in and if the answer makes sense, then 
$f(a,b)$  is the limit. 

If the denominator does vanish but the numerator does not,
the limit automatically {\it does not exist}, end of story.

Unfortunately, in most cases that you are
likely to get, both top and bottom are zero at
the designated point $(a,b)$, so you must go on 
to the next step.
&
{\bf 1.} Both top and bottom vanish when we plug-in $(x,y)=(0,0)$
so we must go on.

\cr

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{\bf 2.} The limit may still exist, but proving it formally
is beyond the scope of this class, so chances are that
the limit does not exist, so try to prove that the
limit does {\bf not} exist by finding limits along
straight lines. $(y-b)=c(x-a)$.

&
{\bf 2.} Plugging $y=cx$ in $f(x,y)$ we get that the
limit as $(x,y) \rightarrow (0,0)$ along the line
$y=cx$ is
$$
\lim_{x \rightarrow 0} {{2x^2} \over {3x^2+4(cx)^2}}=
\lim_{x \rightarrow 0} {{2x^2} \over {(3+4c^2)x^2}}=
$$
$$
\lim_{x \rightarrow 0} {{2} \over {(3+4c^2)}}=
{{2} \over {(3+4c^2)}} \quad.
$$
Since this {\bf depends} on the slope $c$, we get {\bf different}
limits for different lines, hence it is not always the same
hence the {\bf limit} (proper) {\bf does not exist}.

{\bf Ans.:} The limit does not exist since you get different limits
when you approach the point $(0,0)$ on different lines.
\cr}

{\bf Problem Type 14.2b}: Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (a,b)} f(x,y) \quad .
$$

{\bf Example Problem 14.2b}: 
Find the limit if it exists, or show that the
limit does not exist.
$$
\lim_{(x,y) \rightarrow (0,0)} {{2x^2} \over {\sqrt{x^2+y^2}}}
\quad .
$$
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&
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} If the function is {\it continuous} at the
point, which is the case if both top and bottom are
`nice' functions like polynomials or exponentials or
sines or cosines {\bf and} the denominator does not
vanish at the designated point $(a,b)$, then just
plug-it in and if the answer makes sense, then 
$f(a,b)$  is the limit. 

If the denominator does vanish but the numerator does not,
the limit automatically {\it does not exist}, end of story.

Unfortunately, in most cases that you are
likely to get both top and bottom are zero at
the designated point $(a,b)$, so you must go on 
to the next step.
&
{\bf 1.} Both top and bottom vanish when we plug-in $(x,y)=(0,0)$
so we must go on.

\cr

\bigskip & \bigskip \cr
{\bf 2.} The limit may still exist, but proving it formally
is beyond the scope of this class, so chances are that
the limit does not exist, so try to prove that the
limit does {\bf not} exist by finding limits along
straight lines $(y-b)=c(x-a)$, hoping that you
would get an expression that {\bf depends on c},
thereby showing that there are ``differenet limits for
different lines'' and hence the {\bf limit} itself
definitely  {\bf does not exist}.

Alas, if the limit does {\bf not} depend on $c$,
then it is a {\bf plain number}, and that number is
the {\bf candidate limit}.

&
{\bf 2.} Plugging $y=cx$ in $f(x,y)$ we get that the
limit as $(x,y) \rightarrow (0,0)$ along the line
$y=cx$ is
$$
\lim_{x \rightarrow 0} {{2x^2} \over {\sqrt{x^2+(cx)^2}}}
$$
$$
=\lim_{x \rightarrow 0} {{2x^2} \over {(\sqrt{(1+c^2)x^2} )}}
=\lim_{x \rightarrow 0} {{2x^2} \over {(\sqrt{(1+c^2) \vert x \vert} )}}
$$
$$
=\lim_{x \rightarrow 0} {{2x^2} \over {(\sqrt{(1+c^2) \vert x \vert} )}}
=\lim_{x \rightarrow 0} {{2\vert x\vert} \over {(\sqrt{(1+c^2)} )}}=0 \quad.
$$
Since this does {\bf not} depend on the slope $c$,
there is a {\bf good chance} that the limit exists
and is $0$, but this is {\bf not} a proof, since
if you approach the point in question $(0,0)$ via
other paths you might get different values, and then
the limit would not exist.

\cr

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{\bf 3.} Proving formally that the above {\bf candidate limit}
is indeed correct,  using $\epsilon-\delta$,
is beyond the scope of this class. So we have to resort
to a {\it trick}, that often works:
converting to {\it polar coordinates},
$$
x=r \cos \theta \quad y=r \sin \theta \quad .
$$
&
{\bf 3.} The function becomes (since $r^2=x^2+y^2$)
$$
f(r \cos \theta, r \sin \theta)=
{{2(r \cos \theta)^2} \over {r}}=
2r \cos^2 \theta \quad ,
$$
and the limit becomes
$$
\lim_{r \rightarrow 0} 2r \cos^2 \theta \quad .
$$
Now everything is nice and continuous (nothing blows-up)
so just plug-it in $r=0$  and get $2 \cdot 0 \cdot \cos^2 \theta =0$,
and the limit is indeed $0$.

{\bf Ans.:} The limit exists and equals $0$.

{\bf Note:} The {\it origin} is $r=0$ {\it not} $(r, \theta)=(0,0)$ so you
{\bf can't} plug-in $r=0$ {\bf and} $\theta=0$. It might happen that when
you plug-in  $r=0$ and $\theta=0$ you would get $0$ but the limit
still does not exist.
\cr}

{\bf Problem Type 14.2c}: 
Determine the set of points
at which the given function $f(x,y)$ is continuous

{\bf Example Problem 14.2c}: Determine the set of points
at which the funcion is continuous:
$$
(a) f(x,y)={{1} \over {x+y+2}} \quad ,
$$
$$
(b) f(x,y)=\sqrt{x+y-3} \quad ,
$$
$$
(c) f(x,y)={{\ln(x+y)} \over {y-3}} \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} The function is continuous whenever
``it makes sense'' and does not blow-up.
Remember that whenever you have something at the
bottom, you must insist that it is {\bf not zero}
for it not to make trouble. Also $\sqrt{w}$ 
and $\ln w$ makes sense only when $w>0$.
&
{\bf 1.} 
(a) The function is continuous whenever the bottom is
not zero, i.e. $x+y+2 \neq 0$.
{\bf Ans. to (a):} The function is continuous everywhere
{\bf except the line} $x+y+2=0$.

(b) Because of the square-root, the function is 
continuous whenever $x+y-3>0$. 

{\bf Ans. to (b):} The function is continuous in the half place
$x+y>3$.

(c) To make the numerator happy we must insist that
$x+y>0$ and to make the denominator happy we must
demand that $y \neq 3$.

{\bf Ans. to (c)}: The half-plane $x+y>0$ without the line $y=3$.

\cr}

{\bf Problem from a previous Final}

Compute the limit
$$
\lim_{(x,y,z) \rightarrow (1,1,1)} e^{-xy} \sin (\pi z/2) \quad ,
$$
or prove that it does not exist.

{\bf Ans.}: It exists and equals ${{1} \over {e}}$.

\end


differenet
funcion