#### Show the proof

#### Hide 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.

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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.