Proving The Existence of Limits Examples 2

# Proving The Existence of Limits Examples 2

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 $\lim_{x \to a} x = a$.

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

Clearly if we choose $\delta = \epsilon$, then both inequalities hold always, so $\lim_{x \to a} x = a$.

## Example 2

Prove that $\lim_{x \to 0} x^2 = 0$.

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

(1)
\begin{align} \mid x^2 - 0 \mid < \epsilon \\ \mid x^2 \mid < \epsilon \\ x^2 < \epsilon \\ \sqrt{x^2} = \mid x \mid < \sqrt{\epsilon} \end{align}

So let $\delta = \sqrt{\epsilon}$. If $0 < \mid x < \delta$, then:

(2)
\begin{align} \quad \mid x^2 \mid = x^2 = (\sqrt{x^2})^2 = (\mid x \mid)^2 < (\sqrt{\epsilon})^2 \\ \mid x^2 \mid < \epsilon \end{align}

Therefore $\lim_{x \to 0} x^2 = 0$.

## Example 3

Prove that $\lim_{x \to 6} x^2 - 6 = 30$.

Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that if $0 < \mid x - 6 \mid < \delta$ then $\mid (x^2 - 6) - 30 \mid < \epsilon$.

We start with the rightmost inequality to try to derive an appropriate delta, $\mid (x^2 - 6) - 30 \mid = \mid x^2 - 36 \mid < \epsilon$.

(3)
\begin{align} \mid x^2 - 36 \mid < \epsilon \\ \mid (x + 6) (x - 6) \mid < \epsilon \\ \mid x + 6 \mid \mid x - 6 \mid < \epsilon \end{align}

We need to find somewhat of an "upper bound" for $\mid x + 6$. Assume that $\delta ≤ 1$ so that $0 < \mid x - 6 \mid < \delta ≤ 1$. Therefore $\mid x - 6 \mid < 1 \implies -1 < x - 6 < 1 \implies 11 < \mid x + 6 \mid < 13$. Therefore we have an upper bound of 13 which is illustrated below:

Therefore,

(4)
\begin{align} \mid x + 6 \mid \mid x - 6 \mid < 13 \mid x - 6 \mid < \epsilon \\ \quad 13 \mid x - 6 \mid < \epsilon \implies \mid x - 6 \mid < \frac{\epsilon}{13} \end{align}

Now we choose $\delta = \mathrm{min} \{1, \frac{\epsilon}{13} \}$ since we need to show that FOR ALL $\epsilon > 0$ the definition of a limit holds. We note that the definition holds for $\delta ≤ 1$, so it will also hold for any values less than 1 such as $\frac{\epsilon}{13}$. Including this ensures all $\epsilon$ are covered, including large $\epsilon > 13$.

So if $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{13} \}$. First consider the case where $\epsilon < 13$ so then $\delta = \frac{\epsilon}{13}$.

(5)
\begin{align} \quad \mid (x^2 - 6) - 30 \mid = \mid x + 6 \mid \mid x - 6 \mid \\ \mid (x^2 - 6) - 30 \mid < 13 \mid x - 6 \mid \end{align}

But since $0 < \mid x - 6 \mid < \delta = \frac{\epsilon}{13}$:

(6)
\begin{align} \mid (x^2 - 6) - 30 \mid < 13 \mid x - 6 \mid < 13 \cdot \frac{\epsilon}{13} \\ \mid (x^2 - 6) - 30 \mid < \epsilon \end{align}

Now consider the case where $\delta = 1$. We note that $\frac{\epsilon}{13} > 1$ since we chose $\delta$ to be the minimum of those two values. Therefore

(7)
\begin{align} \mid (x^2 - 6) - 30 \mid < \mid x + 6 \mid \mid x - 6 \mid \\ \quad \mid (x^2 - 6) - 30 \mid < 13 \mid x - 6 \mid < 13 \cdot 1 < 13 \cdot \frac{\epsilon}{13} \\ \mid (x^2 - 6) - 30 \mid < \epsilon \end{align}