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

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