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\textbf{Problem 27 - Section 15.3}

As discussed in class, the idea is that we first convert the functions to polar coordinates.
Once converted to polar, we then take the limit as $r \ra 0$.
If this limit does not depend on $\theta$, then the limit 
\[
\lim_{(x,y) \ra (0,0)} f(x,y)
\]
exists and is equal the limit of the polar function as $r \ra 0$.
If the limit as $r \ra 0$ depends on $\theta$ (and we can show this by example), then the limit
will be different for different paths to the origin and will therefor not exist.

I will only write the converted equations below.

\begin{enumerate}[(a)]
\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    r\cos\theta\sin\theta & r \neq 0
  \end{cases}
  \]
  As discussed in class, we can apply the squeeze theorem to this now:
  \[
  -1 \leq \cos\theta\sin\theta \leq 1
  \]
  Therefore, 
  \[
  -r \leq r\cos\theta\sin\theta \leq r
  \]
  Applying the squeeze theorem gives the limit:
  \[
  \lim_{r \ra 0} f(r,\theta) = 0
  \]
  and therefore the function is continuous.

\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    \cos\theta\sin\theta & r \neq 0
  \end{cases}
  \]
  If we take the limit
  \[
  \lim_{r \ra 0} f(r,\theta)
  \]
  we see one can get lots of different values for different values of $\theta$.
  and therefore the function is discontinuous.


\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    r^{1/3}\cos\theta & r \neq 0
  \end{cases}
  \]
  If we take the limit, applying the squeeze theorem as in part (a):
  \[
  \lim_{r \ra 0} f(r,\theta) = 0
  \]
  and therefore the function is continuous.


\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    r^2\cos\theta\sin\theta(\cos^2\theta - \sin^2\theta) & r \neq 0
  \end{cases}
  \]
  If we take the limit, applying the squeeze theorem as in part (a):
  \[
  \lim_{r \ra 0} f(r,\theta) = 0
  \]
  and therefore the function is continuous.


\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    r^2 \sin^2\theta \left( \frac{\cos^2\theta}{\cos^2\theta + r^2 \sin^4 \theta} \right) & r \neq 0
  \end{cases}
  \]
  Consider the following expression:
  \[
  \frac{a^2}{a^2+b^2}
  \]
  Notice that regardless of what $a$ and $b$ are (unless both are zero), 
  the denominator of this expression is always less then (or equal to) the numerator.
  Therefore, the entire fraction will be less then or equal to one:
  \[
  \frac{a^2}{a^2+b^2} \leq 1
  \]
  We now apply this logic to our situation letting $a=\cos\theta$ and $b=r\sin^2\theta$.
  (Notice that we can't have both $a$ and $b$ zero at the same time.)
  This gives:
  \[
  \frac{\cos^2\theta}{\cos^2\theta + r^2 \sin^4 \theta} \leq 1
  \]
  Therefore, we have
  \[
  r^2 \sin^2\theta \left( \frac{\cos^2\theta}{\cos^2\theta + r^2 \sin^4 \theta} \right) \leq r^2 \sin^2\theta
  \leq r^2
  \]
  Taking the limit (and applying the squeeze theorem) gives:
  \[
  \lim_{r \ra 0} f(r,\theta) = 0
  \]
  and therefore the function is continuous.


\item  
  \[
  f(r,\theta) = \begin{cases}
    0 & r=0 \\
    \frac{r\cos\theta\sin^2\theta}{\cos^2\theta + r^2 \sin^4 \theta} & r \neq 0
  \end{cases}
  \]
  Notice that the reasoning of the previous problem doesn't quite work.
  So, we start experimenting.

  If $\sin\theta =0$ ($\theta=0$) then the function simplifies to $f(r,0) = 0$ and therefore the limit 
  $\lim_{r\ra 0} f(r,0)= 0$.

  Similarly, if $\theta=\pi/2$, then $f(r,\pi/2) = 0$ and the limit is $0$ and things are looking OK.

  In fact, for any fixed value of $\theta$, the limit of the function (as $r \ra 0$) will be zero and things are looking great.
  What this means geometrically is that if you approach the origin directly (as in a line, which is what a fixed value of $\theta$ means),
  then the function has a limit of zero. (Can you work this out?)

  But, this doesn't tell us what happens to the function as we approach the origin in a funny way. 
  For example, what happens if we sprial in, rather then approach in a straight line?

  One approach at this point is to start trying to come up with examples of paths to the origin
  (that aren't straight lines) so that the limit of the function is not zero.
  Another approach is to try and prove that regardless of the approach, the limit is always zero 
  (this is what we did in the previous part).
  I think it is generally a good idea to start trying to contruct counter examples 
  (a path where the limit is not zero) to better understand the problem, so this is the approach we take:

  Look at the function and notice the similarity to the expression:
  \[
  \frac{rab^2}{a^2+r^2b^4}
  \]
  (what are $a$ and $b$?)
  The way that we can make this not equal to zero is to make sure that the denominator goes to 
  zero faster then the numerator.
  In order to make sure that the denominator goes to zero is to make sure $a \ra 0$ 
  (in other words, $\cos \theta \ra 0$ and therefore $\theta \ra \pi/2$).
  Notice that if $\cos\theta \ra 0$, then $\sin\theta \ra 1$ (OK, it is possible that $\sin\theta \ra -1$, but we will pretend that this doesn't happen).

  If $\sin \theta \ra 1$, then on our scratch paper (I'm including this though so you can see it), we write:
  \begin{eqnarray}
  f(r,\theta) & = & \frac{r\cos\theta\sin^2\theta}{\cos^2\theta + r^2 \sin^4 \theta} \nonumber \\
  & \approx & \frac{r\cos\theta}{\cos^2\theta + r^2} \label{simple}
  \end{eqnarray}

  Now, we stare at this simplified version of the function and try to come up with a relationship
  between $r$ and $\theta$ so that the limit as $r \ra 0$ is not equal to zero.
  I came up with $r=\cos \theta$.
  In polar coordinates, this is the circle centered at $(1/2,0)$, with radius $1/2$.
  If you believe this (you should graph it), then you can see that as $\theta \ra \pi/2$, then $r \ra 0$.
  We now substitue $r=\cos \theta$ into the equation above (equation number~(\ref{simple})):
  \begin{eqnarray*}
  f(\cos \theta,\theta) & \approx & \frac{\cos\theta \cos\theta}{\cos^2\theta + \cos^2\theta} \\
  & = & \frac{1}{2}
  \end{eqnarray*}
  So, this looks promising.
  What happens if we substitute $r=\cos\theta$ into the original function:
  \begin{eqnarray*}
  f(\cos\theta,\theta) & = & \frac{\cos\theta\cos\theta\sin^2\theta}{\cos^2\theta + \cos^2\theta \sin^4 \theta} \\
  & = &  \frac{\sin^2\theta}{1 + \sin^4 \theta}
  \end{eqnarray*}
  Taking the limit of this function as $\theta \ra \pi/2$ gives $1/2$, which is not equal to $0$.
  Therefore the limit:
  \[
  \lim_{(x,y) \ra (0,0)} f(x,y)
  \]
  cannot exist because we found two path to the origin with different limits.

  I realize that this probably sounds a bit complicated, but let me summarize the main points:
  \begin{enumerate}[1.]
  \item  The main point is to just start trying to take limits of the function 
    along paths to the origin.
  \item  We first try the ``obvious'' paths to the origin: the lines (which are given by equations
    of the form $\theta=\mbox{a constant}$).
    These paths all give limits of zero.
  \item  Now we try some more creative paths to the origin.  The answer book just tells you to
    try going along $x=y^2$, but I don't think it gives any motivation for this.
    I tried to give a bit of motivation for my choice of path ($r=\cos\theta$).
    In either case, you get a limit along this path that is nonzero.
  \item  IF you had been unsuccessful in finding a path that gave a different limit,
    then I would start trying to prove that you always get a limit of zero regardless of the path.
    Sometimes this can be a bit tricky.
  \end{enumerate}
  
\end{enumerate}

\end{document}