Prove that \(\lim\limits_{x \rightarrow 2} (2x + 4)= 8\)

Definitions and Theorems:

Plan

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

$$ \begin{align*} |2x + 4 - 8| < \epsilon \end{align*} $$

We want to choose \(\delta\) which controls how close \(x\) gets \(2\). We want to choose it such that it ensures that \(f(x)\) is close to \(8\). So we can work backward and start from the conclusion to derive what \(\delta\) needs to be.

$$ \begin{align*} |2x + 4 - 8| &< \epsilon \\ |2x - 4| &< \epsilon \\ 2|x - 2| &< \epsilon \\ |x - 2| &< \frac{\epsilon}{2} \end{align*} $$

so as long as we bound \(|x-2|\) by \(\delta = \frac{\epsilon}{2}\), then we can bound the functional limit.

Proof

Let \(\epsilon > 0\) be arbitrary. Let \(\delta = \frac{\epsilon}{2}\). Now, suppose that

$$ \begin{align*} |x - 2| < \delta = \frac{\epsilon}{2} \end{align*} $$

Observe now that

$$ \begin{align*} |2x + 4 - 8| &= |2x - 4| \\ &= 2|x - 2| \\ &< 2 \cdot \frac{\epsilon}{2} = \epsilon \end{align*} $$

Thus, \(|(2x + 4) - 8| < \epsilon\) as desired. \(\blacksquare\)

References

  • Problem Statement Source: Aleph 0