Proving The Existence of Limits Examples 5

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

Let $f(x) = \biggr \{ \begin{matrix} x^2 & \mathrm{if} \: x \in \mathbb{Q}\\ x & \mathrm{if} \: x \not \in \mathbb{Q} \end{matrix}$. Prove that $\lim_{x \to 1} f(x) = 1$.

We now have somewhat of a more difficult problem to overcome since the answer is not so obvious to start with. First, let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that if $0 < \mid x - 1 \mid < \delta$, then $\mid f(x) - 1 \mid < \epsilon$.

First, suppose that $x \in \mathbb{Q}$. Then $f(x) = x^2$, and thus, we need to find $\delta_1 > 0$ such that if $0 < \mid x - 1 \mid < \delta_1$ then $\mid x^2 - 1 \mid < \epsilon$. Using this equality:

(1)

\begin{align} \mid x^2 - 1 \mid < \epsilon \\ \mid x + 1 \mid \mid x - 1 \mid < \epsilon \\ \end{align}

Now suppose $\delta_1 ≤ 1$ (that is we're restricting $x$ to be within $1$ of $x = 1$ or rather:

(2)

\begin{align} \mid x + 1 \mid < \delta_1 ≤ 1 \\ \mid x + 1 \mid < 1 \\ -1 < x + 1 < 1 \\ -3 < x - 1 < -1 \mid x - 1 \mid < 3 \end{align}

We have thus found somewhat of an "upper bound" for $\mid x - 1 \mid$. Therefore:

(3)

\begin{align} \mid x + 1 \mid \mid x - 1 \mid < \epsilon \\ 3 \mid x - 1 \mid < \epsilon \\ \mid x - 1 \mid < \frac{\epsilon}{3} \end{align}

We will let $\delta_1 = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$.

Now suppose that $x \not \in \mathbb{Q}$. Then $f(x) = x$ and therefore we need to find $\delta_2$ such that if $0 < \mid x - 1 \mid < \delta_2$ then $\mid x - 1 \mid < \epsilon$. Thus, we can clearly see choosing $\delta_2 = \epsilon$ will work.

Therefore, we now have $\delta_1$ and $\delta_2$ which give us three choices of $\epsilon$ to pick. We will choose our main delta to be the minimum of all three, or rather $\delta = \mathrm{min} \{ \delta_1, \delta_2 \} = \mathrm{min} \{1, \frac{\epsilon}{3}, \epsilon \}$. We note that if $\epsilon ≥ 3$, then $\delta = 1$, and if $\epsilon < 3$, then $\delta = \frac{\epsilon}{3}$. So really, we have $\delta = \{ 1, \frac{\epsilon}{3} \}$.

Now we will show that $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$ works.

First suppose that $x \in \mathbb{Q}$ and $\epsilon < 3$. Then $\delta = \frac{\epsilon}{3}$, and therefore:

(4)

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

Now suppose that $x \in \mathbb{Q}$ and $\epsilon ≥ 3$. Then $\delta = 1$ and $1 < \frac{\epsilon}{3}$, and therefore:

(5)

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

Now suppose that $x \not \in \mathbb{Q}$ and $\epsilon < 3$. Then $\delta = \frac{\epsilon}{3}$ and therefore:

(6)

\begin{align} \quad \mid x - 1 \mid < \frac{\epsilon}{3} < \epsilon \end{align}

Now suppose that $x \not \in \mathbb{Q}$ and $\epsilon ≥ 3$. Then $\delta = 1$ and $1 < \frac{\epsilon}{3}$ and therefore:

(7)

\begin{align} \mid x - 1 \mid < 1 < \epsilon \end{align}

Therefore $\lim_{x \to 1} f(x) = 1$.