Proving The Existence of Limits Examples 3

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 $\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = 5$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x - 2 \mid < \delta$ then $\biggr \rvert \frac{x^2 + x - 6}{x - 2} - 5 \biggr \rvert < \epsilon$.

We will first start with the following inequality:

(1)

\begin{align} \biggr \rvert \frac{x^2 + x - 6}{x - 2} - 5 \biggr \rvert < \epsilon \\ \biggr \rvert \frac{x^2 + x - 6 - 5(x - 2)}{x - 2} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{x^2 -4x + 4}{x - 2} \biggr \rvert < \epsilon \\ \biggr \rvert \frac{(x - 2)(x - 2)}{x - 2} \biggr \rvert < \epsilon \\ \mid x - 2 \mid < \epsilon \end{align}

Choose $\delta = \epsilon$. If $0 < \mid x - 2 \mid < \delta = \epsilon$, then:

(2)

\begin{align} \biggr \rvert \frac{x^2 + x - 6}{x - 2} - 5 \biggr \rvert = \mid x - 2 \mid < \delta = \epsilon \\ \biggr \rvert \frac{x^2 + x - 6}{x - 2} - 5 \biggr \rvert < \epsilon \end{align}

Therefore $\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = 5$.

Example 2

Prove that $\lim_{x \to a} \cos x = \cos a$.

First let $\epsilon > 0$ be given. We want to find $\delta > 0$ such that if $0 < \mid x - a \mid < \delta$ then $\mid \cos x - \cos x \mid < \epsilon$.

Now consider the function $f(x) = \cos x$. $f$ is continuous on the closed interval $[A, B]$ and differentiable on the open interval $(A, B)$, so by the Mean Value Theorem $\exists C$ such that $A < C < B$ that satisfies:

(3)

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

We note that $-1 ≤ -\sin C ≤ 1$ which is equivalent to saying $\mid -\sin C \mid = \mid \sin C \mid ≤ 1$. Therefore:

(4)

\begin{align} \frac{\mid \cos B - \cos A \mid}{\mid B - A \mid} = \mid \sin C \mid ≤ 1 \\ \frac{\mid \cos B - \cos A \mid}{\mid B - A \mid} ≤ 1 \\ \mid \cos B - \cos A \mid ≤ \mid B - A \mid \end{align}

Now choose $B = x$ and $A = a$. We thus get that $\mid \cos x - \cos a \mid ≤ \mid x - a \mid$ We note that $\mid \cos x - \cos a \mid ≤ \mid x - a \mid < \epsilon$ if we choose $\delta = \epsilon$ since $0 < \mid x - a \mid < \delta$ which forces $\mid \cos x - \cos x \mid < \delta = \epsilon$.

Example 3

Prove that $\lim_{x \to 2} (x^2 - 4x + 5) = 1$.

Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x - 2 \mid < \delta$ then $\mid (x^2 - 4x +5) - 1 \mid < \epsilon$.

We first start off with the following inequality:

(5)

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

We note that if we can find a $C > 0$ such that $\mid x + 2 \mid < C$, then $\mid x + 2 \mid \mid x - 2 \mid < C \mid x - 2 \mid$ and then we can make $C \mid x - 2 \mid < \epsilon$ when $\mid x - 2 \mid < \frac{\epsilon}{C} = \delta$.

Let $x$ be restricted to lie in some interval centered at $2$ We're only interested in values near $2$ so it is reasonable to assume $x$ is within a distance of $1$ from $2$ or in other words, we restrict $\delta ≤ 1 \implies 1 < x < 2$.

So then $2 < x + 1 < 4 \implies 3 < x + 2 < 5 \implies \mid x + 2 \mid < 5$ Therefore our upper bound is $5$.

Therefore $\mid x + 2 \mid \mid x - 2 \mid < 5 \mid x - 2 \mid < \epsilon \implies \mid x - 2 \mid < \frac{\epsilon}{5}$.

We now have two inequalities to satisfy, that is $\mid x - 2 \mid < \delta ≤ 1$ and $\mid x - 2 \mid < \frac{\epsilon}{5}$. To ensure that both inequalities are accounted for, choose $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{3} \}$.

Now consider when $0 < \epsilon < 5$. Then $\delta = \frac{\epsilon}{5}$. So then if $0 < \mid x - 2 \mid < \delta = \frac{\epsilon}{5}$, then:

(6)

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

Now consider when $\epsilon ≥ 5$. Then $\delta = 1$. Now since $\delta$ is defined to be the minimum of two values, this implies that $1 ≤ \frac{\epsilon}{5}$ which we can clearly see if $\epsilon ≥ 5$. Therefore:

(7)

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

Therefore $\lim_{x \to 2} (x^2 - 4x + 5) = 1$.