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\author{Kyle Spaans}
\date{May 8, 2009}
\title{Calculus 3 Lecture Notes}
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\section*{Lecture 3 -- Limits}
Limits in two variables. We have the standard definition from \emph{Calculus 1}
To generalize to more variables, use the formular for Euclidean distance:
\[ \| \vec{x} - \vec{x_0} \| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
So using this in our limit definition:
\[ \forall \epsilon > 0 \, \exists \delta > 0 \, \mathrm{such that} \,
\forall x, y \| \vec{x} - \vec{x_0} \| < \delta
\Rightarrow \mid f(\vec{x}) - L \mid < \epsilon \, \mathrm{and} \,
\vec{x} = (x,y), \vec{x_0} = (x_0, y_0) \]
If you picture this in 2D, we have a circular area around the points $\vec{x}$
and $f(\vec{x})$ of radii $\epsilon$ and $\delta$, respectively. These are called
the neighbourhoods of the points.
When limits don't exist, there are three possible cases:
\begin{itemize}
\item Value goes to infinity ($\displaystyle \frac{1}{x}$ as $x \to 0$)
\item Periodic functions (trigonometric functions)
\item Two different limits from two sides (discontinuities)
\end{itemize}
Some simple examples in two variables:
\paragraph*{Example 1}
\[ \lim_{(x,y) \to (0,0)} \frac{1 + x^2y^6}{3 + x + x^3} = \frac{1}{3} \]
found by direct substitution
\paragraph*{Example 2}
\[ \lim_{(x,y) \to (0,0)} \frac{2 + x + y}{xy^3} = \infty \]
\paragraph*{Example 3}
Cases with $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$.
\[ f(x,y) = \frac{x^2 - y^2}{x^2 + y^2},
\lim_{(x,y) \to (0,0)} \]
We will prove by contradiction that this limit does not exist. Consider first
the case were we get $\displaystyle \frac{0}{0}$. Fix $x = 0$ and let
$y \to 0$, the limit of this is clearly $-1$. Conversely when we fix $y = 0$
and let $x \to 0$ we get a limit of $1$ by l'H\^opital's Rule. Because these
values are different, the limit does not exist.
\paragraph*{Example 4}
\[ f(x,y) = \frac{x^4y^4}{(x^4 + y^2)^3}, \, x,y \neq 0,
\]
When $x = 0$ we get a limit of $0$. Similarly for $y = 0$. So we can't find
a contradiction yet. What if we approach $(0,0)$ from an arbirary line rather
than one of the axes? Let $y = mx$ giving
$\displaystyle \lim_{x \to 0} \frac{x^4(mx)^4}{(x^4 + (mx)^2)}
= \lim_{x \to 0} \frac{m^4x^2}{(x^2 + m^2)^3} = 0$. So we don't know
any more information. We can find a contradition if try try letting $y = mx^2$,
approaching $(0,0)$ from an arbitrary parabola. Solving that equation gives us
$\displaystyle \frac{m^4}{(1 + m^2)^3} \neq 0$, which leads to a condradiction,
hence the limit does not exist.
So we can see that this approach isn't good for finding limits. It is also not
incredibly useful for proving that limits exist as well. We need to figure out
when to stop, and figure out strategies for different approaches to the problem.
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