# Right-Hand Differentiable Function is Right-Continuous

## Theorem

Let $f$ be a real function defined on an interval $I$.

Let $a$ be a point in $I$ where $f$ is right-hand differentiable.

Then $f$ is right-continuous at $a$.

## Proof

By hypothesis, $\map {f'_+} a$ exists.

First we note that $a$ cannot be the right hand end point of $I$ because values in $I$ greater than $a$ need to exist for $\map {f'_+} a$ to exist.

We form the following expression:

$\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$

We need to show that it is defined and to find its value.

We find:

 $\ds$  $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ $\ds$ $=$ $\ds \lim_{x \mathop \to a^+} \paren {\frac {\map f x - \map f a} {x - a} \paren {x - a} }$ where the denominator is unequal to $0$ since $x > a$ $\ds$ $=$ $\ds \lim_{x \mathop \to a^+} \paren {\frac {\map f x - \map f a} {x - a} } \lim_{x \mathop \to a^+} \paren {x - a}$ Product Rule for Limits of Real Functions since (see the next step) the two limits exist $\ds$ $=$ $\ds \map {f'_+} a \times 0$ Definition of Right-Hand Derivative $\ds$ $=$ $\ds 0$

Note that this proves that $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ exists.

We continue by manipulating the result above:

 $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a} + \map f a - \map f a$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a} + \lim_{x \mathop \to a^+} \map f a - \map f a$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a + \map f a} - \map f a$ $=$ $\ds 0$ Sum Rule for Limits of Real Functions since the two limits in the previous expression exist $\ds \leadstoandfrom \ \$ $\ds \lim_{x \mathop \to a^+} \map f x - \map f a$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \lim_{x \mathop \to a^+} \map f x$ $=$ $\ds \map f a$

which means that $f$ is right-continuous at $a$.

$\blacksquare$