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\newtheorem{remark*}[theorem]{Remark}

\title{$\epsilon-\delta$ examples}
\date{}
\begin{document}
\maketitle


\begin{prop}
The function $f(x) = x$ is continuous at $x = 0$.
\end{prop}

\begin{proof}
Given $\epsilon > 0$, we need to show there exists a $\delta > 0$ so that $|x - 0| < \delta$ implies $|f(x)-f(0)| < \epsilon$. Since
%
\begin{align*}
|f(x)-f(0)| = |x-0|, 
\end{align*}
%
we can simply set $\delta = \epsilon$. Then if $|x-0| < \delta$, we have
%
\begin{align*}
|f(x)-f(0)| = |x-0| < \delta = \epsilon.
\end{align*}
%
Therefore, $f$ is continuous at $x =0$.  
\end{proof}


\begin{prop}
The function $\mathbf{x}(t) = (t,t^2)$ is continuous at $t = 2$.   
\end{prop}

{\bf Remark:} It's possible to prove this by first proving more general theorems about continuity, and then deducing the above as a special case. However, we give a direct argument to illustrate the $\epsilon-\delta$ style argument.

\begin{proof}
We need to show that $\lim_{t\to 2} \mathbf{x}(t) = \mathbf{x}(2)$, or in other words for all $\epsilon > 0$ there exists a $\delta > 0$ such that $|t-2| < \delta$ implies $|\mathbf{x}(t) - \mathbf{x}(2)| < \epsilon$. 

First we compute
%
\begin{align*}
|\mathbf{x}(t) - \mathbf{x}(2)|^2 
&= |(t,t^2) - (2,4)|^2 \\
&= (t-2)^2 + (t^2-4)^2 \\
&= (t-2)^2 + ((t-2+2)^2 - 4)^2 \\
&=  (t-2)^2 + ((t-2)^2 + 4(t-2) + 4 - 4)^2 \\
&= (t-2)^2 + ((t-2)^2 + 4(t-2))^2 \\
&= (t-2)^2 + (t-2)^4 + 8(t-2)^3 + 16(t-2)^2,
\end{align*}
%
This might be clearer if we wrote $\mathbf{x}(2+h) - \mathbf{x}(2)$ (try it!), but the idea is that we're going to make $|t-2|$ small, which is why we try to get everything in those terms.

At this point it's pretty clear that we can make the right hand side as small as we want by forcing $|t-2|$ to be small. To be precise, when $|t-2| < 1$, we have that $(t-2)^4,(t-2)^3 < (t-2)^2$. Therefore so long as we make sure that $|t-2| < 1$, we have
%
\begin{align*}
|\mathbf{x}(t) - \mathbf{x}(2)|^2 
&< (t-2)^2 + (t-2)^4 + 8(t-2)^3 + 16(t-2)^2 \\
&< (t-2)^2 + (t-2)^2 + 8(t-2)^2 + 16(t-2)^2 \\
&= 26(t-2)^2. 
\end{align*}
%
hence
%
\begin{align*}
|\mathbf{x}(t) - \mathbf{x}(2)| < \sqrt{26}|t-2|.
\end{align*}
%
If we want to force the left hand side to be less than $\epsilon$, we see that it is enough to force $|t-2|$ to be less than $\frac\epsilon{\sqrt{26}}$. Therefore if we let $\delta = \min(1,\frac\epsilon{\sqrt{26}})$, then we have that 
%
\begin{align*}
|t-2| < \delta \implies |\mathbf{x}(t) - \mathbf{x}(2)| < \sqrt{26}|t-2| < \sqrt{26}\frac\epsilon{\sqrt{26}} = \epsilon.
\end{align*}
%
Therefore $\lim_{t\to 2} \mathbf{x}(t) = \mathbf{x}(2)$, and thus $\mathbf{x}(t)$ is continuous at $t = 2$. 
\end{proof}



\begin{prop*}
Let $\mathbf{x}(t)$ be a parameterized curve in $\mathbf{R}^n$. If there exists a $\mathbf{v} \in \mathbf{R}^n$ and a vector valued function $\mathbf{r}(t)$ such that
%
\begin{align}\label{exp}
\mathbf{x}(t) = \mathbf{x}(t_0) + (t-t_0)\mathbf{v} + \mathbf{r}(t), 
\end{align}
%
and 
%
\begin{align}\label{bigO}
\exists C > 0 \text{ such that } |\mathbf{r}(t)| \leq C|t-t_0|^2,\text{ whenever }|t-t_0| < 1/C, 
\end{align}
%

then $\mathbf{x}(t)$ is differentiable at $t = t_0$, and $\mathbf{x}'(t_0) = \mathbf{v}$. 
\end{prop*}

{\bf Remark:} If the condition (\ref{bigO}) holds,  we say that $\mathbf{r}(t) = \mathcal{O}(|t-t_0|^2)$ as $t\to t_0$.

\begin{proof}
We need to show that 
%
\begin{align*}
\lim_{h\to 0} \frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} = \mathbf{v}, 
\end{align*}
%
or in other words for all $\epsilon > 0$ there exists a $\delta > 0$ so that 
%
\begin{align*}
|h| < \delta \implies \left|\frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{v}\right| < \epsilon.
\end{align*}

Using (\ref{exp}), we may write
%
\begin{align*}
\frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{v} 
&= \frac{(\mathbf{x}(t_0) + h\mathbf{v} + \mathbf{r}(t+h))- \mathbf{x}(t_0)}{h} - \mathbf{v} \\
&= \frac{\mathbf{r}(t+h)}h.
\end{align*}
%
Then by (\ref{bigO}), we have 
%
\begin{align*}
\left| \frac{\mathbf{r}(t+h)}h \right| \leq \frac{C|(t_0+h)-t_0|^2}{|h|} = C|h|,
\end{align*}
%
so long as $|h| < \frac 1C$. We want to force $h$ to be small enough so that the left hand side is less than $\epsilon$. Thus we notice that if $|h|<\frac{\epsilon}{C}$, then $C|h| < \epsilon$. 

Therefore, let $\delta = \min(\frac 1C , \frac \epsilon{C})$, which is greater than 0. If $|h| < \delta$ then because $|h| < \frac 1C$ and $|h| < \frac \epsilon{C}$, we have
%
\begin{align*}
\left| \frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{v} \right| < C|h| < C \frac \epsilon{C} = \epsilon.
\end{align*}
%
Therefore, for any $\epsilon > 0$ we can find a $\delta > 0$ so that $|h| < \delta$ implies 
%
\begin{align*}
\left| \frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{v} \right| < \epsilon.
\end{align*}
%
Therefore,
%
\begin{align*}
\lim_{h\to 0} \frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} = \mathbf{v}. 
\end{align*}

\end{proof}


\begin{prop}
If $\mathbf{x}(t),\mathbf{y}(t)$ are curves in $\mathbf{R}^3$ which are differentiable at $t = t_0$, then the curve $\mathbf{x}(t) \times \mathbf{y}(t)$ is differentiable at $t = t_0$ and 
\begin{align*}
\left.\frac{d}{dt}\right|_{t=t_0} [\mathbf{x}(t) \times \mathbf{y}(t)] = \mathbf{x}'(t_0) \times \mathbf{y}(t_0) + \mathbf{x}(t_0) \times \mathbf{y}'(t_0). 
\end{align*}
\end{prop}

\begin{proof}
For later, note that since $\mathbf{x}(t)$ and $\mathbf{y}(t)$ are differentiable at $t = t_0$, they are also continuous there by one of the homework problems. We need to show that
%
\begin{align*}
\lim_{h\to 0}\frac{ \mathbf{x}(t_0+h) \times \mathbf{y}(t_0+h) - \mathbf{x}(t_0) \times \mathbf{y}(t_0)}h = \mathbf{x}'(t_0) \times \mathbf{y}(t_0) + \mathbf{x}(t_0) \times \mathbf{y}'(t_0)
\end{align*}
or in other words for all $\epsilon > 0$ there exists a $\delta > 0$ such that $|h| < \delta$ implies 
%
\begin{align*}
\left| \frac{ \mathbf{x}(t_0+h) \times \mathbf{y}(t_0+h) - \mathbf{x}(t_0) \times \mathbf{y}(t_0)}h - (\mathbf{x}'(t_0) \times \mathbf{y}(t_0) + \mathbf{x}(t_0) \times \mathbf{y}'(t_0)) \right| < \epsilon.
\end{align*}

After adding and subtracting  $\mathbf{x}(t_0) \times \mathbf{y}(t_0+h)$ in  the middle of the numerator of the fraction, and regrouping, we get
%
\begin{align*}
&\frac{ \mathbf{x}(t_0+h) \times \mathbf{y}(t_0+h) - \mathbf{x}(t_0) \times \mathbf{y}(t_0)}h - (\mathbf{x}'(t_0) \times \mathbf{y}(t_0) + \mathbf{x}(t_0) \times \mathbf{y}'(t_0)) \\
&= \left(\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h}\right) \times \mathbf{y}(t_0+h) + \mathbf{x}(t_0) \times \left(\frac{\mathbf{y}(t_0+h) - \mathbf{y}(t_0)}h \right) - (\mathbf{x}'(t_0) \times \mathbf{y}(t_0) + \mathbf{x}(t_0) \times \mathbf{y}'(t_0)).
\end{align*}
Adding and subtracting $\mathbf{x}'(t_0) \times \mathbf{y}(t_0+h)$ and $\left(\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{x'}(t_0)\right) \times \mathbf{y}(t_0)$ and regrouping again, we get
%
\begin{align*}
\left(\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{x'}(t_0)\right) \times (\mathbf{y}(t_0+h) -\mathbf{y}(t_0)) 
+\left(\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{x'}(t_0)\right) \times \mathbf{y}(t_0) \\
+\mathbf{x}'(t_0) \times (\mathbf{y}(t_0+h) - \mathbf{y}(t_0)) 
+\mathbf{x}(t_0) \times \left(\frac{\mathbf{y}(t_0+h) - \mathbf{y}(t_0)}h - \mathbf{y}'(t_0)\right).
\end{align*}
%
The point of the above manipulations is that since $\mathbf{x}$ and $\mathbf{y}$ are differentiable at $t_0$, and $\mathbf{y}$ is continuous at $t_0$, by forcing $|h|$ to be small enough we can force each of the terms
%
\begin{align*}
\frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}h - \mathbf{x}'(t_0),\quad
\frac{\mathbf{y}(t_0+h) - \mathbf{y}(t_0)}h - \mathbf{y}'(t_0),\quad
\mathbf{y}(t_0+h) - \mathbf{y}(t_0)
\end{align*}
to be as small as we like. Between these, we expect to be able to make the above quantity as small as desired. More precisely, by the inequality $|\mathbf{u} \times \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|$, the size of the big expression above is at most
%
\begin{align*}
\left|\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{x'}(t_0)\right||\mathbf{y}(t_0+h) -\mathbf{y}(t_0)| 
+\left|\frac{ \mathbf{x}(t_0+h) - \mathbf{x}(t_0)}{h} - \mathbf{x'}(t_0)\right||\mathbf{y}(t_0)|\\
+|\mathbf{x}'(t_0)||\mathbf{y}(t_0+h) - \mathbf{y}(t_0)| 
+|\mathbf{x}(t_0)|\left|\frac{\mathbf{y}(t_0+h) - \mathbf{y}(t_0)}h - \mathbf{y}'(t_0)\right|. 
\end{align*}
%
Let us suppose that 
%
\begin{align}\label{want}
&\left|\frac{\mathbf{x}(t_0+h) - \mathbf{x}(t_0)}h - \mathbf{x}'(t_0)\right| < \epsilon_1\\
&\left|\frac{\mathbf{y}(t_0+h) - \mathbf{y}(t_0)}h - \mathbf{y}'(t_0)\right| < \epsilon_2\\ 
&|\mathbf{y}(t_0+h) - \mathbf{y}(t_0)| < \epsilon_3. 
\end{align}
%
(From what was said before, we know that for any $\epsilon_1,\epsilon_2,\epsilon_3 > 0$, we can find $\delta_1,\delta_2,\delta_3 > 0$ such that $|h| < \delta = \min(\delta_1,\delta_2,\delta_3)$ implies that these are all true.) Then the quantity above is at most
%
\begin{align*}
\epsilon_1\epsilon_2+\epsilon_1|\mathbf{y}(t_0)|+|\mathbf{x}'(t_0)|\epsilon_3+|\mathbf{x}(t_0)|\epsilon_2.
\end{align*}
%
So given $\epsilon > 0$, let 
%
\begin{align*}
\epsilon_1 &= \frac{\epsilon}{4(|\mathbf{y}(t_0)| + 1)} > 0 \\
\epsilon_2 &= \min\left(\frac{\epsilon}{4(|\mathbf{x}(t_0)| + 1)},\frac{\epsilon}{4\epsilon_1}\right) > 0 \\
\epsilon_3 &= \frac{\epsilon}{4(|\mathbf{x}'(t_0)| + 1)}> 0. \\
\end{align*}
%
(The +1's are there just to ensure that the quantities in the denominators are nonzero). Find $\delta_1,\delta_2,\delta_3$ so that $|h| < \delta = \min(\delta_1,\delta_2,\delta_3)$ implies (\ref{want}) is true. Then if $|h| < \delta$, we have that the big expression from before is at most
%
\begin{align*}
&\epsilon_1\epsilon_2+\epsilon_1|\mathbf{y}(t_0)|+|\mathbf{x}'(t_0)|\epsilon_3+|\mathbf{x}(t_0)|\epsilon_2 \\
&< \epsilon_1 \frac{\epsilon}{4\epsilon_1}
+\frac{\epsilon}{4(|\mathbf{y}(t_0)| + 1)}|\mathbf{y}(t_0)|
+|\mathbf{x}'(t_0)|\frac{\epsilon}{4(|\mathbf{x}'(t_0)| + 1)}
+|\mathbf{x}(t_0)|\frac{\epsilon}{4(|\mathbf{x}(t_0)| + 1)} \\
&< \frac{\epsilon}4+\frac{\epsilon}4+\frac{\epsilon}4+\frac{\epsilon}4 = \epsilon.
\end{align*}
%
This concludes the proof.

\end{proof}

\end{document}




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