Continuity of the Derivative of a Function

Continuity of the Derivative 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:

\begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \end{align}

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:
\begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{align}
  • 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:
\begin{align} \quad f(x) - f(c) &= f(x) - f(c) \\ \quad f(x) - f(c) &= \frac{f(x) - f(c)}{x - c} (x - c) \end{align}
  • Taking the limit as $x \to c$ gives us that:
\begin{align} \quad \lim_{x \to c} [f(x) - f(c)] &= \lim_{x \to c} \left ( \frac{f(x) - f(c)}{x - c} (x - c) \right ) \\ \quad \lim_{x \to c} [f(x) - f(c)] &= f'(c) \cdot 0 \\ \quad \lim_{x \to c} [f(x) - f(c)] &= 0 \\ \quad \lim_{x \to c} f(x) &= f(c) \end{align}
  • 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:

\begin{align} \quad f(x) = \mid x \mid \end{align}

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

\begin{align} \quad f'(0) &= \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \\ \quad f'(0) &= \lim_{h \to 0} \frac{\mid h \mid}{h} \end{align}

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 $0$.

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