Proving the Existence of Limits at Infinity

Recall from the Limits at Infinity page that we say that:

• $\lim_{x \to \infty} f(x) = L$ means that $\forall \epsilon > 0 \: \exists k \in \mathbb{R} \: \mathrm{s.t.} \forall x : x > k, \: \mid f(x) - L \mid < \epsilon$.

• $\lim_{x \to -\infty} f(x) = L$ means that $\forall \epsilon > 0 \: \exists k \in \mathbb{R} \: \mathrm{s.t.} \forall x : x < k \: \mid f(x) - L \mid < \epsilon$.

Example 1

Prove that $\lim_{x \to \infty} \frac{1}{x} = 0$.

Let $\epsilon > 0$ be given and find $k \in \mathbb{R}$ such that if $x > k$ then $\mid \frac{1}{x} - 0 \mid < \epsilon$.

We will first start off with the following inequality:

(1)

\begin{align} \biggr \rvert \frac{1}{x} \biggr \rvert < \epsilon \end{align}

Now assume $k > 0$ so that we can get rid of the absolute value bars, then:

(2)

\begin{align} \frac{1}{x} < \epsilon \\ x > \frac{1}{\epsilon} \\ \end{align}

Now let $k = \mathrm{max} \{ 0, \frac{1}{\epsilon} \} = \frac{1}{\epsilon}$. We note that $k$ will never be 0 since $\epsilon > 0$. Therefore:

(3)

\begin{align} \quad x > k =\frac{1}{\epsilon} \implies x > \frac{1}{\epsilon} \implies \frac{1}{x} < \epsilon \implies \biggr \rvert \frac{1}{x} \biggr \rvert < \epsilon \end{align}

Example 2

Prove that $\lim_{x \to \infty} \sqrt{x} = 0$.

Note that is this example, we will have to somewhat "combine" the definitions of a what it means for a limit at infinity that also goes to infinity.

What we really want to so is that $\forall m > 0 \: \exists k \in \mathbb{R} \: \mathrm{s.t.} \forall x : x > k, \: f(x) > m$.

So let $m > 0$ be given. We want to find a $k \in \mathbb{R}$ such that if $x > k$ then $\sqrt{x} > m$. Start with the following inequality:

(4)

\begin{align} \sqrt{x} > m \implies x > m^2 \end{align}

So choose $k = m ^2$, so then:

(5)

\begin{align} x > k = m^2 \implies x > m^2 \implies \sqrt{x} > m \end{align}

Therefore $\lim_{x \to \infty} \sqrt{x} = \infty$.