Week 6: $\epsilon - \delta$ Definition of Limit

            As someone who has no prior experience with the $\epsilon - \delta$ definition of limit, I will teach myself about them with the help of my favourite math textbook and some creativity.  This is what I gleaned so far.

            In high school, we calculate limits.  In university, we like to think that we are sophisticated thinkers. Knowing how to calculate is not enough for us.  We must know why our calculation is correct.  Then we need a definition to work with.  This is where the the $\epsilon - \delta$ definition of limit comes in:

            Let $f:X \to R$ be a function, $L \in R$, $a \in X$ and $X \subseteq R$.  Then

$$\lim_{x \to a} f(x) = L \iff$$

$$\forall \epsilon \in R^+, \exists \delta \in R^+, (\forall x \in R, |x - a| < \delta \implies |f(x) - L| < \epsilon).$$

After digesting this definition, we can prove or disprove statements about the limits of functions.

For example, consider the following statement:

$$\lim_{x \to 2} x^2 = 4$$

By substituting two into $x$, we discover that this statement should be true.  To show that our hunch is correct, we plug in $a = 2$ and $L = 4$ into the $\epsilon - \delta$ definition and prove it:

$$\forall \epsilon \in R^+, \exists \delta \in R^+, (\forall x \in R, |x - 2| < \delta \implies |x^2 – 4| < \epsilon)$$

Our value for $\epsilon$ will be arbitrary but our value for $\delta$ must be concrete since $\epsilon$ has a $\forall$ quantifier but $\delta$ has a $\exists$ quantifier.  The hard part is choosing $\delta$.  How do we do it?

Where to Start

            Think of $\epsilon$ as a distance from $L$ on the vertical axis and $\delta$ as a distance from $2$ on the horizontal axis.  Once the distance of $x$ from $2$ is less than $\delta$ on the horizontal, then the $\epsilon – \delta$ definition of limit guarantees that the distance of $x^2$ from $4$ is less than $\epsilon$ on the vertical.  Hence, our choice of $\delta$ must satisfy the definition for all $\epsilon$.  To achieve this, we express $\delta$ in terms of $\epsilon$, which is done by working backwards from the definition.

How to Work Backwards

       Since the definition is an implication, then working backwards begins with the consequent:

$|x^2 - 4| < \epsilon$

Since we operate in $R$, then we can manipulate the left side of the inequality:

$|x + 2||x - 2| = |x^2 - 4| < \epsilon$

We see that the $|x - 2|$ in $|x + 2||x - 2| < \epsilon$ will be useful in getting the antecedent of the definition, which is $|x - 2| < \delta$.  However, there is still a bit of baggage to unload.

The Creative Step

            Since we choose our $\delta$, then we can give it an upper bound.  For simplicity, we choose $1$ as an upper bound for $\delta$.  So if $|x - 2| < d$, then $|x - 2| < 1$.

            From $|x - 2| < 1$, the absolute value definition gives $-1 < x - 2 < 1$.  By adding $4$ to all sides of the inequality, then we have $3 < x + 2 < 5$.  Use the absolute value definition again, which gives us $|x + 2| < 5$.

Now we can unload $|x + 2|$ from $|x + 2||x - 2| < \epsilon$.  Since $|x + 2| < 5$, then $|x + 2||x - 2| < 5|x - 2|$.  Then $5|x - 2| < e$ is equivalent to $|x - 2| < \frac{\epsilon}{5}$.

Endgame

We can now chose $\delta$.  If $|x - 2| < \frac{\epsilon}{5}$ is true, then its equivalent statement $5|x - 2| < e$ guarantees that $|x - 2||x + 2| = |x^2 - 4| < \epsilon$ since $|x - 2| < 5$.  Yet we chose $1$ as an upper bound for $\delta$.  So if $\frac{\epsilon}{5}>1$, then $\delta = 1$.  Hence our choice of $\delta$ is $min(1, \frac{\epsilon}{5})$.  The proof follows naturally from the scratch work so it will be omitted.