# How do you use the epsilon delta definition to prove a limit exists?

Aug 16, 2015

It depends on the example, but basically you show

"for any $\epsilon > 0 \exists \delta > 0 : \ldots$"

in order to prove

"$\forall \epsilon > 0 \exists \delta > 0 : \ldots$"

See an example in explanation.

#### Explanation:

Consider the function

$q \left(x\right) = \left\{\left(1 , \text{if x = 0"), (1/q, "if x = p/q for integers p, q in lowest terms"), (0, "if x is irrational}\right)\right.$

Then $q \left(x\right)$ is continuous at every irrational number and discontinuous at every rational number.

Let us show that $q \left(x\right)$ is continuous at any irrational number $\alpha$ by showing that ${\lim}_{x \to \alpha} q \left(x\right)$ exists and is zero.

Let $\alpha$ be an irrational number and $\epsilon > 0$

We need to show that:

$\exists \delta > 0 : \forall x \in \left(\alpha - \delta , \alpha + \delta\right) , \left\mid q \left(x\right) - 0 \right\mid < \epsilon$

Let $I = \left(\alpha - 1 , \alpha + 1\right)$ so $\alpha \in I$

Let $N = \left\lceil \frac{1}{\epsilon} \right\rceil + 1$, so $\frac{1}{N} < \epsilon$.

Let $S = \left\{\frac{p}{q} : p , q \in \mathbb{Z} , 1 \le q \le N , \text{hcf} \left(p , q\right) = 1\right\} \cap I$

Then $S$ is finite and for all $x \in S$ we have $\left\mid x - \alpha \right\mid > 0$ since $\alpha$ is irrational.

Note that if $x \in I \mathmr{and} x \notin S$ then $\left\mid q \left(x\right) \right\mid < \frac{1}{N}$

Let $\delta = {\min}_{x \in S} \left\mid x - \alpha \right\mid$

Then since $S$ is finite, $\delta > 0$.

From our definition of $\delta$, if $x \in \left(\alpha - \delta , \alpha + \delta\right)$, then $x \in I$, $x \notin S$ and $\left\mid q \left(x\right) \right\mid < \frac{1}{N}$.

So $\forall x \in \left(\alpha - \delta , \alpha + \delta\right) , \left\mid q \left(x\right) - 0 \right\mid < \frac{1}{N} < \epsilon$