A Simple Case
Let's start with
f(x)=2x
.
We want to prove
\lim_{x \to 1} f(x) = 2
.
In terms of the Epsilon/Delta Method:
\forall \epsilon \gt 0, \exists \delta \gt 0
where if (|x-1| \lt \delta)
, then (|f(x)-2| \lt \epsilon)
. We start by defining
\delta
in terms of
\epsilon
. In this case,
\delta = \epsilon / 4
.
Thus, if
\epsilon = 1, \delta = \frac{1}{4}
. If
\epsilon = \frac{1}{50}, \delta = \frac{1}{200}
.
Now that we know
\delta
, we will prove that when
(|x - 1| \lt \delta)
,
|f(x)-2| \lt \epsilon
.
\begin{align*}
\forall x, \, \text{ where } |x - 1| &\lt \delta \\ \text{} \\
|x - 1| &\lt \frac{\epsilon}{4} \\ \text{} \\
2 * |x - 1| &\lt \frac{\epsilon}{2} \\ \text{} \\
|2x - 2| &\lt \frac{\epsilon}{2} \\ \text{} \\
|f(x) - 2| &\lt \frac{\epsilon}{2} \\ \text{} \\
|f(x) - 2| &\lt \epsilon
\end{align*}
We have proven that
\lim_{x \to 1} f(x) = 2
.
A Less Simple Case
Let's start with
f(x)=x^2
.
We want to prove
\lim_{x \to 0} f(x) = 0
.
In terms of the Epsilon/Delta Method:
\begin{align*}
\delta &= \sqrt{\epsilon} \\ \text{} \\
\forall x, \, \text{ where } |x - 0| &\lt \delta \\ \text{} \\
|x - 0| &\lt \sqrt{\epsilon} \\ \text{} \\
|x - 0|^2 &\lt \sqrt{\epsilon}^2 \\ \text{} \\
x^2 &\lt \epsilon \\ \text{} \\
f(x) &\lt \epsilon \\ \text{} \\
\text{Since } f(x) \gt 0, \, |f(x)| &\lt \epsilon \\ \text{} \\
|f(x) - 0| &\lt \epsilon
\end{align*}
We have proven that
\lim_{x \to 0} f(x) = 0
.
There are much more complex forms of the epsilon/delta proof, and they can get quite challenging to solve. However, that knowledge is not needed right now.
We do not need to know how to write Epsilon/Delta proofs, but we should know how they work.