Proving the Limit at a Point of a Function

# Proving the Limit at a Point of a Function

Recall from The Limit of a Function page that if $f : A \to \mathbb{R}$ and $c$ is a cluster point of $A$ then $\lim_{x \to c} f(x) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. We will now use this formal definition of a limit of a function at a point $c$ in actual proofs.

## Example 1

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2 + 3$. Prove that $\lim_{x \to 1} x^2 + 3 = 4$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $x \in \mathbb{R}$ and $0 < \mid x - 1 \mid < \delta$ then $\mid x^2 + 3 - 4 \mid < \epsilon$.

Looking on the latter inequality, notice that $\mid x^2 + 3 - 4 \mid = \mid x^2 - 1 \mid = \mid (x + 1)(x - 1) \mid = \mid x + 1 \mid \mid x - 1 \mid$. We ultimately want this inequality to be less than $\epsilon$. To do so, we want to keep the term "$\mid x - 1 \mid$" and try to find a constant to replace $\mid x + 1 \mid$.

Suppose that $\delta ≤ 1$. Then we have that $\mid x - 1 \mid < \delta ≤ 1$ and so $\mid x - 1 \mid < 1$ which implies that $-1 < x - 1 < 1$ and so $1 < x +1 < 3$. Therefore $\mid x + 1 \mid < 3$.

So then $\mid x + 1 \mid \mid x - 1 \mid < 3 \mid x - 1 \mid$ and $3 \mid x - 1 \mid < \epsilon$ if $\mid x - 1 \mid < \frac{\epsilon}{3}$.

Therefore, let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$. This will ensure us that both $\mid x - 1 \mid < 1$ and $\mid x - 1 \mid < \frac{\epsilon}{3}$. Thus we have found a $\delta$ (dependent on $\epsilon$) that satisfies the conditions. Therefore $\lim_{x \to 1} x^2 + 3 = 4$.

## Example 2

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2 - 4$. Prove that $\lim_{x \to 4} x^2 - 4 = 12$.

Let $\epsilon > 0$ be given. We want to find $\delta > 0$ such that if $x \in \mathbb{R}$ and $0 < \mid x - 4 \mid < \delta$ then $\mid x^2 - 4 - 12 \mid < \epsilon$.

Notice from the latter inequality that $\mid x^2 - 4 - 12 \mid = \mid x^2 - 16 \mid = \mid (x + 4)(x - 4) \mid$. We want to be less than $\epsilon$ by keeping the term "$\mid x - 4 \mid$" and finding a constant to replace "$\mid x + 4 \mid$."

Assume $\delta ≤ 1$. Then $\mid x - 4 \mid < \delta ≤ 1$, and so $\mid x - 4 \mid < 1$ which implies that $-1 < x - 4 < 1$ and so $3 < x < 5$, and also $7 < x + 4 < 9$. Therefore $\mid x + 4 \mid < 9$.

So then $\mid x + 4 \mid \mid x - 4 \mid < 9 \mid x - 4$ and $9 \mid x - 4 \mid < \epsilon$ if $\mid x - 4 \mid < \frac{\epsilon}{9}$.

Therefore we want to let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{9} \}$ so that both $\mid x - 4 \mid < 1$ and $\mid x - 4 \mid < \frac{\epsilon}{9}$. So we have found a $\delta$ (dependent on $\epsilon$ that works, and so $\lim_{x \to 4} x^2 - 4 = 12$.

## Example 3

Let $f : (-\infty, -3) \cup (-3, \infty) \to \mathbb{R}$ be defined by $f(x) = \frac{2}{x + 3}$. Prove that $\lim_{x \to 3} \frac{2}{x + 3} = \frac{1}{3}$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $x \in (-\infty, 3) \cup (-3, \infty)$ and $0 < \mid x - 3 \mid < \delta$ then $\biggr \rvert \frac{2}{x + 3} - \frac{1}{3} \biggr \rvert < \epsilon$.

Looking at the latter inequality, notice that:

(1)
\begin{align} \quad \quad \biggr \rvert \frac{2}{x + 3} - \frac{1}{3} \biggr \rvert = \biggr \rvert \frac{6}{3(x + 3)} - \frac{(x + 3)}{3(x+3)} \biggr \rvert = \biggr \rvert \frac{6 - (x + 3)}{3 (x + 3)} \biggr \rvert = \biggr \rvert \frac{-(x - 3)}{3(x + 3)} \biggr \rvert = \frac{\mid x - 3 \mid}{3 \mid x + 3 \mid} \end{align}

Assume that $\delta ≤ 1$. Then $\mid x - 3 \mid < \delta ≤ 1$ and so $\mid x - 3 \mid < 1$ which implies that $-1 < x - 3 < 1$ and so $5 < x + 3 < 7$, and so $\frac{1}{7} < \frac{1}{x + 3} < \frac{1}{5}$ and finally $\frac{1}{\mid x + 3 \mid} < \frac{1}{5}$.

Therefore we have that $\frac{\mid x - 3 \mid}{3 \mid x + 3 \mid} < \frac{\mid x - 3 \mid}{3 \cdot 5} = \frac{\mid x - 3 \mid}{15}$ and $\frac{\mid x - 3 \mid}{15} < \epsilon$ if $\mid x - 3 \mid < 15 \epsilon$.

So let $\delta = \mathrm{min} \{ 1, 15 \epsilon \}$ and so we have found a $\delta$ (dependent on $\epsilon$ that works, and so $\lim_{x \to 3} \frac{2}{x + 3} = \frac{1}{3}$.