Theorem 1. If a function $$f$$ is differentiable at $$x_{0}$$ (i.e. if $$f'(x_{0})$$ exists), then $$f$$ is continuous at $$x=x_{0}$$.

#### Show the proof

To show $$f$$ is continuous at $$x=x_{0}$$, we need to show:
(1) $$f(x_{0})$$ exists
(2) $$\lim_{x\to x_{0}}f(x)$$ exists and
(3) $$\lim_{x\to x_{0}}f(x)=f(x_{0})$$.

According to the hypothesis $$f'(x_{0})$$ exists; that is,
$\lim_{\Delta x\to0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}\tag{a}$ 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 $$[f(x_{0}+\Delta x)-f(x_{0})]/\Delta x$$ cannot tend to a finite limit unless $\lim_{\Delta x\to0}[f(x_{0}+\Delta x)-f(x_{0})]=0$ or, equivalently $\lim_{\Delta x\to0}f(x_{0}+\Delta x)=f(x_{0}).$ This means that $$f$$ is continuous at $$x=x_{0}$$.

The above theorem tells us that

$\bbox[#F2F2F2,5px,border:2px solid black] {\text{Differentiability}\Rightarrow\text{Continuity}}$

• It follows from the above theorem that if a function $$f$$ is discontinuous at a point $$x_{0}$$, then $$f'(x_{0})$$ does not exist.

But:

Continuity does not imply differentiability.

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.