Proving The Existence of Limits

# Proving The Existence of Limits

Recall from the precise definition of a limit on the Introduction to Limits page, we said that the statement $\lim_{x \to a} f(x) = L$ says that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < \mid x - a \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. We will now actually look at proving limits exist with the precise definition of the limit, which is sometimes a tricky task.

## Example 1

Prove that $\lim_{x \to 1} x^2 = 1$.

To show that this limit exists, we want to show that $\forall \epsilon > 0 \: \exists \delta > 0 \: \mathrm{s.t.} \forall x : 0 < \mid x - 1 \mid < \delta \: , \: \mid x^2 - 1 \mid < \epsilon$. We should note that we have no control over $\epsilon$. However, we must find a suitable $\delta > 0$ dependent on $\epsilon$ such that if the first inequality holds, then the second inequality holds. If we can do that, then we will have proven that $\lim_{x \to 1} x^2 = 1$.

We will first start with the following inequality and build a list of equivalent inequalities which hold just by using basic algebraic manipulation.

(1)
\begin{align} \mid f(x) - L \mid < \epsilon \\ \implies \mid x^2 - 1 \mid < \epsilon \\ \implies \mid (x + 1)(x - 1) \mid < \epsilon \\ \implies \mid x + 1 \mid \mid x - 1 \mid< \epsilon \\ \end{align}

Now we know how $\mid x - 1 \mid$ behaves as $x \to 1$ from our inequality that $0 < \mid x - 1 \mid < \delta$, but we don't necessarily know how $\mid x + 1\mid$ behaves. In essence, we want to find an upper bound for $\mid x + 1 \mid$. Now suppose the we have the scenario such that $\delta ≤ 1$. We note that we're making an assumption here, so we'll have to account for it later as well. We thus have that:

(2)
\begin{align} \mid x - 1 \mid < \delta ≤ 1 \\ \implies \mid x - 1 \mid < 1 \\ \implies -1 < x - 1 < 1 \\ \implies 0 < x < 2 \\ \implies 1 < \mid x + 1 \mid < 3 \end{align}

The following diagram illustrates that if $\mid x - 1 \mid < \delta ≤ 1$, then the greatest value $x$ can take over the interval $(a - \delta, a + \delta) = (1 - \delta, 1 + \delta) = (0, 2)$ is $3$.

So at most $\mid x + 1 \mid$ is going to be almost equal to 3 given that $\mid x + 1 \mid < \epsilon$, and at minimum it will be equal to 1, and therefore, applying this rule to our inequality we obtain that:

(3)
\begin{align} \: \mid x + 1 \mid \mid x - 1 \mid < 3 \mid x - 1 \mid < \epsilon \\ \implies 3 \mid x - 1 \mid < \epsilon \\ \implies \mid x - 1 \mid < \frac{\epsilon}{3} \end{align}

So we will now choose $\delta = \mathrm{min} \{ 1, \frac{1}{3} \epsilon \}$. We do this step for a very important reason. We supposed that $\delta ≤ 1$, so if we just let $\frac{\epsilon}{3} = \delta ≤ 1$, then what if $\epsilon > 3$? Then we have that $\delta > 1$, and inequalities will no longer hold. Remember by the definition of a limit, we must show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that… If $\epsilon > 3$ though, then there does not exist a $\delta > 0$. Hence, we will choose $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$. If $\epsilon ≤ 3$, then $\delta = \frac{\epsilon}{3}$ and everything is fine, and if $\delta > 3$, then $\delta = 1$ which also works. Note that this $\delta$ may not necessarily be the most "efficient". In other words, there may be a larger $\delta$ to which the inequalities hold. However, we note that any small $\delta$ will also be adequate regardless since if a large interval satisfies the inequalities, then a smaller subset of that interval will also satisfy the inequalities.

Now given $\epsilon > 0$, let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$. If $0 < \mid x - 1 \mid < \delta$, then:

(4)
\begin{align} \quad \mid x^2 - 1 \mid = \mid x + 1 \mid \mid x - 1 \mid \\ \implies \mid x^2 - 1 \mid < 3 \mid x - 1 \mid \mathrm{\left [Since \: \delta = min \{1, \frac{\epsilon}{3} \}.\right]} \end{align}

Now we note that if $0 < \mid x - 1 \mid < \delta = \frac{\epsilon}{3}$, then clearly $\mid x - 1 \mid < \frac{\epsilon}{3}$, so then:

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

Furthermore, if $\delta = 1$, then $0 < \mid x - 1 \mid < \delta = 1$. But we denoted $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$ which implies that $1$ was smaller than $\frac{\epsilon}{3}$, so then:

(6)
\begin{align} \quad \mid x^2 -1 \mid < \delta = 1 < \frac{\epsilon}{3} \\ \quad \mid x^2 - 1 \mid < 3 \mid x - 1 \mid < \frac{\epsilon}{3} \\ \quad \mid x^2 - 1 \mid < \epsilon \end{align}

Therefore if $0 < \mid x - 1 \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. We have shown that $\lim_{x \to 1} x^2 = 1$.

## Example 2

Prove that $\lim_{x \to a} \sin x = \sin a$ (Hint: Use the Mean Value Theorem):

To prove that this limit exists, we must show that $\forall \epsilon > 0 \: \exists \delta > 0 \: \mathrm{s.t.} \: \forall x: 0 < \mid x - a \mid < \delta, \: \mid \sin x -\sin a \mid < \epsilon$.

We will start with the right inequality. We somehow have to introduce "$\mid x - a \mid$". We will do this using the Mean Value Theorem. We first consider the function $f(x) = \sin x$ that is continuous on the closed interval $[A, B]$ and differentiable on the open interval $(A, B)$. By the Mean Value Theorem, $\exists C \: \mathrm{s.t.} A < C < B$ and that satisfies:

(7)
\begin{align} \frac{f(B) - f(A)}{B - A} = f'(C) \\ \frac{\sin B - \sin A}{B - A} = \cos C \end{align}

We now essentially want to "get rid" of $C$ with a formal argument. We know that $-1 ≤ \cos C ≤ 1$ so then $\mid \cos C \mid ≤ 1$, and therefore:

(8)
\begin{align} \biggr \rvert \frac{\sin B - \sin A}{B - A} \biggr \rvert = \mid \cos C \mid ≤ 1 \\ \implies \biggr \rvert \frac{\sin B - \sin A}{B - A} \biggr \rvert ≤ 1 \\ \implies \mid \sin B - \sin A \mid ≤ \mid B - A \mid \end{align}

Now suppose that we let $B = x$ and $A = a$ so thus we obtain that:

(9)
\begin{align} \mid \sin x - \sin a \mid ≤ \mid x - a \mid \end{align}

Now if we let $\delta = \epsilon$, then $\mid \sin x - \sin a \mid < \mid x - a \mid < \epsilon$. We know this is true because $0 < \mid x < \delta$ which forces $\mid \sin x - \sin a \mid < \delta = \epsilon$.