Limits of Functions: Example 2

Example 2: Prove that $\lim\limits_{x \rightarrow 2} x^2 = 4$

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

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

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 $4$. To do so, we want to work backward. That it, what would we need to set delta to in order to have $|x^2 - 4| < \epsilon$. So we can start from the conclusion and work our way backwards.

$$ \begin{align*} |x^2 - 4| &< \epsilon \\ |(x-2)(x+2)| &< \epsilon \\ |x-2||x+2| &< \epsilon \end{align*} $$

We want to bound $|x-2|$. So what do we do with $|x+2|$? We know we can control the gap between $x$ and $2$ with $\delta$. That is in fact $\delta$’s definition. Okay, so if we assume that $\delta < 1$, meaning that $|x - 2| < \delta < 1$. Then

$$ \begin{align*} |x - 2| &< 1 \\ |x| - 2 &< 1 \\ |x| &< 3 \end{align*} $$

So if we set this arbitrarily chosen limit on $\delta$, then it implies that $|x| < 3$. But this means that

$$ \begin{align*} |x + 2| &< 5 \end{align*} $$

So now

$$ \begin{align*} |x-2||x+2| &< \epsilon \\ |x-2| \cdot 5 &< \epsilon \\ |x-2| &< \frac{\epsilon}{5} \end{align*} $$

so as long as we bound $|x-2|$ by $\delta = \frac{\epsilon}{5}$, then we can bound the functional limit.

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

Problem Statement Source: Aleph 0