Proving The Existence of Limits Examples 4

# Proving The Existence of Limits Examples 4

We will now look at some more examples of proving the existence of limits using the definition of a limit, that is $\lim_{x \to a} f(x) = L$ says that $\forall \epsilon > 0 \: \exists \delta > 0 \: \mathrm{s.t.} \: \forall x : 0 < \mid x - a \mid < \delta, \mid f(x) - L \mid < \epsilon$. More examples can be found on the following pages:

## Example 1

Prove that $\lim_{x \to 0} \mid x \mid = 0$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x \mid < \delta$ then $\mid \mid x \mid \mid < \epsilon$.

We note that $\mid \mid x \mid \mid = \mid x \mid < \epsilon$. So choose $\delta = \epsilon$ so that if $0 < \mid x \mid < \delta = \epsilon$ then $\mid \mid x \mid \mid < \epsilon$.

Therefore $\lim_{x \to 0} \mid x \mid = 0$.

## Example 2

Prove that $\lim_{x \to -2} (x^2 - 1) = 3$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x + 2 \mid < \delta$ then $\mid (x^2 - 1) - 3 \mid < \epsilon$.

(1)
\begin{align} \mid (x^2 - 1) - 3 \mid < \epsilon \\ \mid x^2 - 4 \mid < \epsilon \\ \mid x - 2 \mid \mid x + 2 \mid < \epsilon \end{align}

We now want to find a $C > 0$ such that $\mid x - 2 \mid < C$. We will restrict $x$ to lie in some interval centered at $-2$ and within a distance of $1$ or in other words, $\delta ≤ 1$. Therefore:

(2)
\begin{align} \mid x + 2 \mid < \delta ≤ 1 \\ \mid x + 2 \mid < 1 \\ -1 < x + 2 < 1 \\ -5 < x - 2 < -3 \\ \mid x - 2 \mid < 5 \end{align}

Therefore we choose $C = 5$ and thus:

(3)
\begin{align} \: \mid x - 2 \mid \mid x + 2 \mid < 5 \mid x + 2 \mid < \epsilon \\ \mid x + 2 \mid < \frac{\epsilon}{5} \end{align}

Now we need to choose a $\delta$ that satisfies $\mid x + 2 \mid < \delta ≤ 1$ and $\mid x + 2 \mid < \frac{\epsilon}{5}$. We will choose $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{5} \}$.

Now suppose that $\epsilon < 5$. Then $\delta = \frac{\epsilon}{5}$, and so:

(4)
\begin{align} \quad \mid (x^2 - 1) - 3 \mid = \mid x^2 - 4 \mid = \mid x -2 \mid \mid x + 2 \mid < 5 \mid x + 2 \mid < 5 \cdot \frac{\epsilon}{5} \\ \mid (x^2 - 1) - 3 \mid < \epsilon \end{align}

Now suppose that $\epsilon ≥ 5$. Then $\delta = 1$. Since $\delta$ is defined to be the minimum of $1$ and $\frac{\epsilon}{5}$, it follows that $1 ≤ \frac{\epsilon}{5}$, and so:

(5)
\begin{align} \quad \mid (x^2 - 1) -3 \mid = \mid x - 2 \mid \mid x + 2 \mid < 5 \mid x + 2 \mid < 5 \cdot 1 ≤ \epsilon \\ \mid (x^2 - 1) - 3 \mid < \epsilon \end{align}

Therefore $\lim_{x \to -2} (x^2 - 1) = 3$.

## Example 3

Prove that $\lim_{x \to 2} x^3 = 8$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x - 2 \mid < \delta$ then $\mid x^3 - 8 \mid < \epsilon$.

(6)
\begin{align} \mid x^3 - 8 \mid < \epsilon \\ \mid (x^2 + 2x + 4)(x - 2) \mid < \epsilon \\ \: \mid x^2 + 2x + 4 \mid \mid x - 2 \mid < \epsilon \end{align}

We want to find a $C > 0$ such that $\mid x^2 + 2x + 4 \mid < C$. Suppose that we restrict $x$ to be within 1 from 2. That is, $\delta ≤ 1$. Therefore $\mid x^2 + 2x + 4 \mid < 19$ as shown with the following diagram:

We can also solve this algebraically. If $\delta ≤ 1$, then:

(7)
\begin{align} \mid x - 2 \mid < \delta ≤ 1 \\ \mid x - 2 \mid < 1 \\ -1 < x - 2 < 1 \\ -(x - 2) < (x - 2)^2 < (x - 2) \\ -x + 2 < x^2 -4x + 4 < x - 2\\ 5x + 2 < x^2 + 2x + 4 < 7x - 2 \end{align}

Remember we restricted $x$ such that $1 < x < 3$, so then $7 < x^2 + 2x + 4 < 19$. Therefore, we will choose $C = 19$ so then:

(8)
\begin{align} \: \mid x^2 + 2x + 4 \mid \mid x - 2 \mid < 19 \mid x - 2 \mid < \epsilon \\ 19 \mid x - 2 \mid < \epsilon \\ \mid x - 2 \mid < \frac{\epsilon}{19} \end{align}

We will let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{19} \}$.

Now if $\epsilon < 19$, then $\delta = \frac{\epsilon}{19}$ and thus:

(9)
\begin{align} \quad \mid x^3 - 8 \mid = \mid (x^2 + 2x + 4) \mid \mid x - 2 \mid < 19 \mid x - 2 \mid < 19 \cdot \frac{\epsilon}{19} \\ \mid x^3 - 8 \mid < \epsilon \end{align}

Now if $\epsilon ≥ 19$, then $\delta = 1$. But since $\delta$ is the minimum of $1$ and $\frac{\epsilon}{19}$, then this implies that $1 < \frac{\epsilon}{19}$ and thus:

(10)
\begin{align} \quad \mid x^3 - 8 \mid = \mid (x^2 + 2x + 4) \mid \mid x - 2 \mid < 19 \mid x - 2 \mid < 19 \cdot 1 ≤ \epsilon \\ \mid x^3 - 8 \mid < \epsilon \end{align}

Therefore $\lim_{x \to 2} x^3 = 8$.