5.4 Differentiability and Continuity of Functions

To show $f$ is continuous at $x=x_0$, we need to show:
(1) $f(x_0)$ exists
(2) $\underset{x\to x_0}{lim}f(x)$ exists and
(3) $\underset{x\to x_0}{lim}f(x)=f(x_0)$.

According to the hypothesis $f^{\prime }(x_0)$ exists; that is,
$$\begin{array}{}\text{(a)}& \underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x_0+\mathrm{\Delta }x)-f(x_0)}{\mathrm{\Delta }x}\end{array}$$ exists. Therefore, we conclude $f(x_0)$ exists (otherwise (a) is meaningless). This means condition (1) holds true.

The denominator of (a) tends to zero. It is thus evident that $[f(x_0+\mathrm{\Delta }x)-f(x_0)]/\mathrm{\Delta }x$ cannot tend to a finite limit unless $$\underset{\mathrm{\Delta }x\to 0}{lim}[f(x_0+\mathrm{\Delta }x)-f(x_0)]=0$$ or, equivalently $$\underset{\mathrm{\Delta }x\to 0}{lim}f(x_0+\mathrm{\Delta }x)=f(x_0).$$ This means that $f$ is continuous at $x=x_0$.

It is natural to ask whether the converse of  Theorem 1 is true, i.e. whether every continuous curve has a definite tangent at every point, and every function has a derivative for every value of $x$ for which it is continuous.

With an example, we can show that the converse is not true; that is, if a function is continuous at a point, it is not necessarily differentiable there. For instance, in Section on the Derivative Concept, we saw that $f(x)=5|x|-9$ is not differentiable (= its derivative does not exist) at $x=0$ (Figure 1). In general, if the graph of a function has a corner, it cannot have a tangent, and the function is not differentiable there. We will come back to this topic in Section 5.6.