Example 2
Prove that \(\lim\limits_{x \rightarrow 2} x^2 = 4\)
Strategy
Let \(\epsilon > 0\), then we want to show that there exists a \(\delta > 0\) such that when
$$
\begin{align*}
|x - 2| < \delta
\end{align*}
$$
then we have
$$
\begin{align*}
|x^2 - 4| < \epsilon
\end{align*}
$$
Same as before. We can choose \(\delta\) to control how close \(x\) is to \(2\). We want to choose it such that it ensures that \(f(x)\) is close to \(4\). We work backwards as follows:
$$
\begin{align*}
|x^2 - 4| &< \epsilon \\
|(x-2)(x+2)| &< \epsilon \\
|x-2||x+2| &< \epsilon
\end{align*}
$$
\(|x-2|\) is bounded by \(\delta\) but what do we do with \(|x+2|\)? If we let \(\delta = 1\), then
$$
\begin{align*}
|x-2| &< \delta = 1 \\
|x| - |2| &< |x - 2| < 1 \quad \text{(reverse triangle inequality)} \\
|x| &< 1 + |2| = 3.
\end{align*}
$$
But this means that
$$
\begin{align*}
|x + 2| &\leq |x| + |2| \\
|x + 2| &\leq 3 + |2| = 5
\end{align*}
$$
So we’re back to
$$
\begin{align*}
|x-2||x+2| &< \epsilon \\
\delta \cdot 5 &< \epsilon \\
\delta &< \frac{\epsilon}{5}
\end{align*}
$$
So we need to set \(\delta\) to be the minmum of these two bounds: \(1\) and \(\epsilon/5\).
Proof
Let \(\epsilon > 0\) be arbitrary. Let \(\delta = \min\{1,\frac{\epsilon}{5}\}\). Now, suppose that
$$
\begin{align*}
|x - 2| < \delta
\end{align*}
$$
Then since \(|x-2| < 1\), then \(x \in (1,3)\) and so this means that
$$
\begin{align*}
|x + 2| < 5
\end{align*}
$$
Observe now that
$$
\begin{align*}
|x^2 - 4| &= |x+2| \cdot |x-2| \\
&< |x+2| \cdot \frac{\epsilon}{5} \\
&< 5 \cdot \frac{\epsilon}{2} = \epsilon
\end{align*}
$$
Thus, \(|x^2 - 4| < \epsilon\) as desired. \(\blacksquare\)
References