# Continuity and Differentiability of a Function

Recall from The Derivative of a Function page that if $f$ is a function defined on the open interval $(a, b)$ and if $c \in (a, b)$ then $f$ is said to be differentiable at $c \in (a, b)$ if the following limit exist:

(1)The limit $f'(c)$ is called the derivative of $f$ at $c$ and the process for which $f'$ is obtained from $f$ is called differentiation.

We will now look at a nice theorem which tells us that if $f$ is differentiable at a point, then $f$ is also continuous at that point.

Theorem 1: Let $f$ be a function defined on the open interval $(a, b)$ and let $c \in (a, b)$. If $f$ is differentiable at $c$ then $f$ is continuous at $c$. |

**Proof:**Suppose that $f$ is differentiable at $c$. Then the following limit exists:

- To show that $f$ is continuous at $c$ we must show that $\displaystyle{\lim_{x \to c} f(x) = f(c)}$. Now we have that:

- Taking the limit as $x \to c$ gives us that:

- So $f$ is continuous at $c$.

It is very important to note that the converse to Theorem 1 is not true in general. There exists certain functions that are continuous at a point but are not differentiable at that point. For example, consider the following function $f : \mathbb{R} \to \mathbb{R}$ defined by:

(5)Consider the point $0 \in \mathbb{R}$. Then:

(6)Now notice that $\displaystyle{\lim_{h \to 0^+} \frac{\mid h \mid}{h} = 1}$ while $\displaystyle{\lim_{h \to 0^-} \frac{\mid h \mid}{h} = -1}$. Since the lefthand limit and righthand limit of $f$ at $0$ are not equal, we have that the $f'(0)$ does not exist, so $f$ is not differentiable at $0$.