Example 1
Prove that \(\lim\limits_{x \rightarrow 2} 3x + 1 = 7\)
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
$$
\begin{align*}
|(3x + 1) - 7| < \epsilon
\end{align*}
$$
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 \(7\). So we can start from the conclusion and work our way backwards as follows
$$
\begin{align*}
|(3x + 1) - 7| &< \epsilon \\
|3x - 6| &< \epsilon \\
3|x - 2| &< \epsilon \\
|x - 2| &< \frac{\epsilon}{3}
\end{align*}
$$
This means in order to achieve the conclusion that we want, the gap between \(x\) and \(2\) needs to be less than \(\frac{\epsilon}{3}\). So we set \(\delta\) to this value.
Proof
Let \(\epsilon > 0\) be arbitrary. Let \(\delta = \frac{\epsilon}{3}\). Suppose that
$$
\begin{align*}
|x - 2| < \delta = \frac{\epsilon}{3}
\end{align*}
$$
Then
$$
\begin{align*}
|x - 2| &< \frac{\epsilon}{3} \\
3|x - 2| &< \epsilon \\
|3x - 6| &< \epsilon \\
|(3x + 1) - 7| &< \epsilon
\end{align*}
$$
But this is exactly \(|f(x) - 7|< \epsilon\) which is what we wanted to show. \(\blacksquare\)
References
- Problem Statement Source: Aleph 0