In this post, we will learn about the epsilon-delta definition of continuity with solved examples. To learn this, let us first recall the definition of continuity.

**Definition of Continuity:**

A real-valued function f(x) is said to be continuous at a point x=a in the domain of f(x) if the following condition is satisfied:

$\lim\limits_{x \to a} f(x)=f(a)$.

## Epsilon Delta Definition of Continuity

A real-valued function f(x) is said to be continuous at x=a if we have the following: For every ε>0, there exists a δ>0 such that

|f(x)-f(a)| < ε whenever 0<|x-a|<δ

holds true.

Let us understand the above ε-δ definition with the help of an example.

**Example:** By the epsilon-delta method, show that $f(x)=x^2 $ is continuous at x=a.

*Solution:*

To show $f(x)=x^2$ is continuous at $x=a$, we need to show the following. For a given positive number ε>0, we need to find a δ>0 such that

$|f(x)-f(a)| < \epsilon$ whenever 0<|x-0|<δ …(I)

that is,

$|x^2-a^2|< \epsilon$ whenever 0<|x-0|<δ …(II)

Choose δ = $\dfrac{\epsilon}{x+a}$. Then we have δ>0. Now, for 0< |x-a| < δ = $\dfrac{\epsilon}{x+a}$ we get that

|(x-a)(x+a)| < |x+a| $\times \dfrac{\epsilon}{x+a}$

$\Rightarrow |x^2-a^2| < \epsilon$ using the formula $(a-b)(a+b)=a^2-b^2$.

Thus we have shown that $|f(x)-f(a)| < \epsilon$ whenever 0 < |x-a| < δ = $\dfrac{\epsilon}{x+a}$. In other words, Equation (I) is true. Hence by epsilon-delta definition of limit, we can conclude that the function $f(x)=x^2$ is continuous at the point $x=a$.

**Note** that the function $f(x)=x^2$ is continuous everywhere.

**Also Read:**

**Epsilon Delta definition of limit**

We will now give the negative statement of the epsilon-delta definition of limit.

## Epsilon Delta Definition of Discontinuity

**Negation of Epsilon Delta Definition of Continuity:** Let f(x) be a continuous function at x=a, that is, $\lim\limits_{x \to a}$ f(x) = f(a)$. The negation of the epsilon-delta definition of continuity is given as follows:

f(x) does not tend to f(a) when x→a then it is not true that for every ε>0, there exists a δ>0 such that

|f(x)-f(a)| < ε, whenever 0<|x-a|<δ.

**That is,** if f(x) does not tend to L when x→a, then there exists some ε>0 such that for every δ>0 there exists some point x in the set {x : 0<|x-a|<δ} such that

|f(x)-f(a)| ≥ ε.

## FAQs

**Q1: Show that f(x)=x is using ε-δ method.**

Answer: Choosing ε=δ in the definition of ε-δ method of continuity, one can easily show that f(x)=x is continuous.