4.2: The Derivative Requires Continuity

Suppose $u$ is a function.

$$\text{If} \quad \lim _{b \rightarrow 3} \frac{u(b) − 5}{b − 3} = 4, \quad \text{what is} \quad \lim _{b \rightarrow 3} u(b) \text{?}$$

The answer is that

$$\lim _{b \rightarrow 3} u(b) = 5$$

We reason that

$$\begin{array}\ \text{for b close to 3, the numerator of} \quad \frac{u(b) − 5}{b − 3} \quad \text{is close to 4 times the denominator.}\\ \text{That is,} \quad u(b) − 5 \quad \text{is close to} \quad 4 \times (b − 3). \end{array}$$

But $4 \times (b − 3)$ is also close to zero. Therefore

If $b$ is close to 3, $u(b) - 5$ is close to zero and $u(b)$ is close to 5.

The general question we address is:

Theorem $\PageIndex{1}$ The Derivative Requires Continuity

If $u$ is a function and $u ^{\prime} (t)$ exists at $t = a$ then $u$ is continuous at $t = a$.

Proof

In Exercise 4.2.2 you are asked to give reasons for the following steps, (i) − (v). Suppose the hypothesis of Theorem 4.2.1.

$$\begin{array}\ \left(\lim _{b \rightarrow a} u(b)\right)-u(a) &=\lim _{b \rightarrow a}(u(b)-u(a)) &(i)\\ &=\lim _{b \rightarrow a} \frac{u(b)-u(a)}{b-a} \times(b-a) &\\ &=\lim _{b \rightarrow a} \frac{u(b)-u(a)}{b-a} \times \lim _{b \rightarrow a}(b-a) &(ii)\\ &=u^{\prime}(a) \lim _{b \rightarrow a}(b-a) &(iii)\\ &=0 &(iv)\\ \left(\lim _{b \rightarrow a} u(b)\right) &=u(a) &(v) \end{array} \label{4.5}$$

End of proof.

A graph of a function $u$ defined by

$$u(t)= \begin{cases}0 & \text { for } \quad 20 \leq t<28 \\ 50 & \text { for } \quad t=28 \\ 100 & \text { for } \quad 28<t \leq 30\end{cases} \label{4.6}$$

is shown in Figure $\PageIndex{1}$A. We observed in Section 4.1 that $u$ is not continuous at $t = 28$. $( \lim _{t \rightarrow 28^−} u(t) = 0 \neq 50 = u(28).)$ Furthermore, $u ^{\prime} (28)$ does not exist. A secant to the graph through $(b, 0)$ and $(28, 50)$ with $b < 28$ is drawn in Figure $\PageIndex{1}$B, and

$$\text { for } b<28, \quad \frac{u(b)-u(28)}{b-28}=\frac{0-50}{b-28}=\frac{50}{28-b}$$

The slope of the secant gets greater and greater as $b$ gets close to 28.

Figure $\PageIndex{1}$: A. Graph of The function $u$ defined in Equation $\ref{4.6}$. B. Graph of $u$ and a secant to the graph through $(b, 0)$ and $(28, 50)$.

Explore 4.2.1 Is there a line tangent to the graph of $u$ shown in Figure $\PageIndex{1}$ at the point (28,50) of the graph?

Figure $\PageIndex{2}$: a. Graph of $P(t) = |t|$ for all $t$. B. Graph of $(P(t) − P(0))/(t − 0)$.

Explore 4.2.2 In Explore Figure 4.2.2 is the graph of $y = \sqrt{|x|}$. Does the graph have a tangent at (0,0)? Your vote counts.

Explore Figure 4.2.2 Graph of $y = \sqrt{|x|}$.

The graph of $P(t) = |t|$ for all $t$ is shown in Figure $\PageIndex{2}$A. $P$ is continuous, but $P ^{\prime} (0)$ does not exist.

$$\frac{P(b)-P(0)}{b-0}=\frac{|b|-0}{b-0}=\frac{|b|}{b}=\left\{\begin{array}{rll} -1 & \text { for } & b<0 \\ 1 & \text { for } & b>0 \end{array}\right.$$

A graph of $\frac{P(b) − P(0)}{b − 0}$ is shown in Figure $\PageIndex{2}$B. It should be clear that

$$\lim _{b \rightarrow 0} \frac{P(b) − P(0)}{b − 0} \quad \text{does not exist, so that} \quad P ^{\prime} (0) \quad \text{does not exist.}$$

Therefore, the converse of Theorem 4.2.1 is not true. Continuity does not imply that the derivative exists.

 

Exercises for Section 4.2, The Derivative Requires Continuity.

Exercise 4.2.1 Shown in Figure 4.2.1 is the graph of $C(t) = \sqrt[3] {t}$.

1. Use Definition of Derivative Equation 3.3.3 to show that $C ^{\prime} (0)$ does not exist.
2. Is $C(t)$ continuous?
3. Is there a line tangent to the graph of $C$ at (0,0)?

Figure for Exercise 4.2.1 Graph of \(C(t) = \sqrt[3] {t} for Exercise 4.2.1.

Exercise 4.2.2 Justify the steps (i) − (v) in Equations $\ref{4.5}$.

Exercise 4.2.3 For for the function $P(t) = |t|$ for all $t$ compute $P ^{\prime -} (0)$ and $P ^{\prime +} (0)$.

Explore 4.2.3 This problem may require extensive thought. Is there a function defined for all numbers $t$ and continuous at every number $t$ and for which $f ^{\prime -} (1)$ does not exist?