Differentiability and Continuity

5.3.1 Theorem. If a function f is differentiable at x₀, then f is continuous at x₀.

Proof. Since f is differentiable at x₀, f'(x₀) exists and is given by

$$f^{\prime }(x_0)=\underset{h\to 0}{lim}\frac{f(x_0+h)-f(x)}{h}$$

Continuity of f at x₀ implies that lim_{h ⟶ 0} f(x) = f(x₀) or, equivalently,

Expressing this in terms of the variable h = x − x₀, and multiplying and dividing by h we have

$$\begin{array}{rl}\underset{h\to 0}{lim}f(x_0+h)-f(x_0)& =\underset{h\to 0}{lim}\frac{f(x_0+h)-f(x)}{h}\cdot h\\ & =\underset{h\to 0}{lim}\frac{f(x_0+h)-f(x)}{h}\cdot \underset{h\to 0}{lim}h\\ & =f^{\prime }(x_0)\cdot 0=0\end{array}$$

therefore lim_{h ⟶ 0} f(x) = f(x₀) and f is continuous at x₀.  □

As a consequence of the previous theorem, discontinuous functions cannot be differentiable. The converse of the theorem is not true, as we shall se in the next section: differentiability is a stronger property than continuity.

Example 5.3.2. The derivative of a function f at x₀ exists if and only if both the left-hand derivative at x₀ and the rigt-hand derivative at x₀ exists and are equal. If f is differentiable at x₀, then f is continuous at x₀. In order to show that a function f is not differentiable at x₀, we can show that the function f is not continuous at x₀, or alternatevely that the left/right-hand derivative at x₀ do not exist.

For example, the piecewise defined function

$$y(x)=\{\begin{array}{ll}{x}^3+2& x<0\\ x^2-1& 0\le x\end{array}$$

is defined on the interval (−∞, ∞). It is continuous on the intervals (−∞, 0) and (0, ∞), but it is not continuous at x = 0, since

Because y(x) is not continuous at x = 0, the function y(x) is not differentiable at x = 0; however, y(x) is differentiable on (−∞, 0) and (0, ∞). In fact,

$$y^{\prime }(x)=\{\begin{array}{ll}3x^2& x<0\\ 2x& 0\le x\end{array}$$

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Corner Points

A continuous function need not to be differentiable, so the converse of Theorem 5.3.1, does not hold. The absolute value function |x| is probably the best known example of a continuous function that is not differentiable.

As we have remarked, if f is differentiable at x, then the right-sided and left-sided derivatives must exist and be equal. In the case of absolute value function, f(x) = |x|, f(x) = x for x > 0 and f(x) = −x for x < 0; We have f'(x) = +1 for x > 0 and f'(x) = −1 for x < 0. To calculate the derivative at the origin we use the definition of derivative

thus, if h ⟶ 0⁺, |h| = h and the limit is 1, while if h ⟶ 0⁻, |h| = −h and the limit is −1.
Since the limit of the increment ratio does not exist, f is not differentiable at x = 0. From the graph of |x| we recognize that at the origin the tangent line is not well-defined, we call this point a corner point. At a corner point, the slopes of the secant lines have different limits from the left and from the right, and hence the two-sided limit that defines the derivative does not exist.

However, the absolute value function is differentiable on (−∞, 0) and (0, ∞), for it coincides with the line y = x when x > 0 and with y = −x for x < 0. Its derivative is

$$|x|=\text{sgn}(x)=\{\begin{array}{ll}1,& \text{if }x\ge 0\\ -1,& \text{if }x<0.\end{array}$$

Vertical Tangents

At a point of vertical tangency the slopes of the secant lines approach +∞ or −∞ from the left and from the right, so again the limit that defines the derivative does not exist.

This is the case for the cube root f(x) = x^1/3 so that f'(x) = 1/(3x^2/3). This function is continuous for all x; it derivative tends to +∞ for x ⟶ 0. We can say that its graph has a vertical tangent at x = 0: the line x = 0 (the y - axis). Note that in the case of |x| the two tangent lines were different, here it's the same. We say that the graph of x^1/3 has a vertical tagent at x = 0.

Cusps

When the graph of a function comes to a sharp point, we say that point on the graph is a Cusp. A cusp, can be traced without lifting your pencil, so the function is continuous at a corner or cusp.

Where f' approached +∞ from one side of x₀ and −∞ from the other side, we say that the graph has a cusp at x₀. For example the function f= x^2/3 has a cursp at x = 0. Differently from |x|, which has two limits "that disagree" that disqualifies it from having a tangent line there, in the case of x^2/3 the tangent line is the same.

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