Proving The Existence of Limits Examples 1

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:

• Proving The Existence of Limits Examples 1
• Proving The Existence of Limits Examples 2
• Proving The Existence of Limits Examples 3
• Proving The Existence of Limits Examples 4
• Proving The Existence of Limits Examples 5

Example 1

Prove that $\lim_{x \to 1} (1 - 4x) = 13$.

First let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x + 3 \mid < \delta$ then $\mid (1 - 4x) - 13 \mid < \epsilon$.

We will start with the following inequality:

(1)

\begin{align} \mid (1 - 4x) - 13 \mid < \epsilon \\ \mid -4x - 12 \mid < \epsilon \\ \mid -4(x + 3) \mid < \epsilon \\ 4 \mid x + 3 \mid < \epsilon \mid x +3 \mid < \frac{\epsilon}{4} \end{align}

We will choose $\delta = \frac{\epsilon}{4}$.

So if $0 < \mid x + 3 \mid < \delta = \frac{\epsilon}{4}$, then:

(2)

\begin{align} \mid (1 - 4x - 13) \mid = 4 \mid x + 3 \mid < 4 \delta = 4 \frac{\epsilon}{4} \\ \mid (1 - 4x - 13) \mid < \epsilon \end{align}

Therefore, $\lim_{x \to -3} (1 - 4x) = 13$.

Example 2

Prove that $\lim_{x \to 1} \frac{2 + 4x}{3} = 2$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x - 1 \mid < \delta$, then $\biggr \rvert \frac{2 + 4x}{3} \biggr \rvert < \epsilon$.

Start with the following inequality:

(3)

\begin{align} \biggr \rvert \frac{2 + 4x}{3} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{2 + 4x - 6}{3} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{4x - 4}{3} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{4(x -1)}{4} \biggr \rvert < \epsilon \\ \frac{4}{3} \mid x - 1 \mid < \epsilon \\ \mid x - 1 \mid < \frac{3}{4} \epsilon \end{align}

Therefore choose $\delta = \frac{3}{4} \epsilon$. If $0 < \mid x - 1 \mid < \delta = \frac{3}{4} \epsilon$, then:

(4)

\begin{align} \quad \biggr \rvert \frac{2 + 4x}{3} \biggr \rvert = \frac{4}{3} \mid x - 1 \mid < \frac{4}{3} \delta = \frac{4}{3} \cdot \frac{3}{4} \epsilon = \epsilon \\ \biggr \rvert \frac{2 + 4x}{3} \biggr \rvert < \epsilon \end{align}

Therefore $\lim_{x \to 1} \frac{2 + 4x}{3} = 2$.