Limits of Functions: Example

Example: Prove that $\lim\limits_{x \rightarrow 2} 3x + 1 = 7$

Plan

Following the definition, 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 $7$. To do so, we want to work backward. That it, what would we need to set delta to in order to have $|(3x + 1) - 7| < \epsilon$. So we can start from the conclusion and work our way backwards.

$$ \begin{align*} |(3x + 1) - 7| &< \epsilon \\ |3x - 6| &< \epsilon \\ 3|x - 2| &< \epsilon \\ |x - 2| &< \frac{\epsilon}{3} \end{align*} $$

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