Proving the Existence of Limits to Infinity

# Proving the Existence of Limits to Infinity

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

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

## Example 1

Prove that $\lim_{x \to 0} \frac{1}{x^2} = \infty$.

By the definition of a limit going to infinity, we get that $\forall k > 0 \: \exists \delta > 0 \: \mathrm{s.t.} \: \forall x : 0 < \mid x - a \mid < \delta, \: f(x) > k$.

Let $k > 0$ be given and find $\delta > 0$ such that if $0 < \mid x \mid< \delta$ then $\frac{1}{x^2} > k$. We will start off with the following inequality:

(1)
\begin{align} \frac{1}{x^2} > k \\ x^2 < \frac{1}{k} \\ \sqrt(x^2) = \mid x \mid < \frac{1}{\sqrt{k}} \end{align}

So choose $\delta = \frac{1}{\sqrt{k}}$. If $0 < \mid x \mid < \delta = \frac{1}{\sqrt{k}}$ then:

(2)
\begin{align} \quad \mid x \mid = \sqrt{x^2} < \frac{1}{\sqrt{k}} \implies \frac{1}{\sqrt{x^2}} > \sqrt{k} \implies \frac{1}{x^2} > k \end{align}

Therefore, $\lim_{x \to 0} \frac{1}{x^2} = \infty$.

## Example 2

Prove that $\lim_{x \to 0} -\frac{1}{x^2} = -\infty$.

By the definition of a limit going to negative infinity, $\forall k < 0 \: \exists \delta > 0 \: \mathrm{s.t.} \: \forall x : 0 < \mid x - a \mid < \delta, \: f(x) < k$.

Let $k < 0$ be given and find $\delta > 0$ such that if $0 < \mid x \mid < \delta$ then $-\frac{1}{x^2} < k$. We will first start off with the following inequality:

(3)
\begin{align} -\frac{1}{x^2} < k \\ -x^2 > \frac{1}{k} \\ x^2 < \frac{1}{-k} \\ \sqrt{x^2} < \frac{1}{\sqrt{-k}} \\ \mid x \mid < \frac{1}{\sqrt{-k}} \end{align}
 Note: Since $k < 0$, then $-k > 0$, so our square root is still defined.

So choose $\delta = \frac{1}{\sqrt{-k}}$. If $0 < \mid x \mid < \delta = \frac{1}{\sqrt{-k}}$, then:

(4)
\begin{align} \quad \mid x \mid = \sqrt{x^2} < \delta = \sqrt{1}{\sqrt{-k}} \implies \sqrt{x^2} < \frac{1}{\sqrt{-k}} \implies x^2 < \frac{1}{-k} \implies -x^2 > \frac{1}{k} \implies -\frac{1}{x^2} < k \end{align}

Therefore $\lim_{x \to 0} -\frac{1}{x^2} = -\infty$.