Differentiability implies continuity

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We see that if a function is differentiable at a point, then it must be continuous at that point.

There are connections between continuity and differentiability.

Differentiability Implies Continuity If $f$ is a differentiable function at $x = a$ , then $f$ is continuous at $x=a$ .

To explain why this is true, we are going to use the following definition of the derivative $f'(a) = \lim _{x\to a} \frac {f(x)-f(a)}{x-a}.$

Assuming that $f'(a)$ exists, we want to show that $f(x)$ is continuous at $x=a$ , hence we must show that $\lim _{x\to a} f(x) = f(a).$ Starting with $\lim _{x\to a} \left (f(x) - f(a)\right )$ we multiply and divide by $(x-a)$ to get

$\begin{align*} &= \lim_{x\to a} \left((x-a)\frac{f(x) - f(a)}{x-a}\right) \\ &= \left(\lim_{x\to a} (x-a) \right) \left(\lim_{x\to a}\frac{f(x) - f(a)}{x-a}\right) &\text{Limit Law.} \\ &= \answer[given]{0}\cdot f'(a) = \answer[given]{0}. \end{align*}$ Since $\lim _{x\to a}\left (f(x) - f(a)\right ) = 0 ,$ we apply the Difference Law to the left hand side $\lim _{x\to a}f(x) - \lim _{x\to a}f(a) = 0 ,$ and use continuity of a constant to obtain that $\lim _{x\to a}f(x) - f(a) = 0 .$ Next, we add $f(a)$ on both sides and get that $\lim _{x\to a}f(x) = f(a).$ Now we see that $\lim _{x\to a} f(x) = f(a)$ , and so $f$ is continuous at $x=a$ .

This theorem is often written as its contrapositive:

If $f(x)$ is not continuous at $x=a$ , then $f(x)$ is not differentiable at $x=a$ .

Thus from the theorem above, we see that all differentiable functions on $\mathbb R$ are continuous on $\mathbb R$ . Nevertheless there are continuous functions on $\mathbb R$ that are not differentiable on $\mathbb R$ .

Which of the following functions are continuous but not differentiable on $\mathbb R$ ?

$x^2$ $\lfloor x \rfloor$ $|x|$ $\frac {\sin (x)}{x}$

Consider $f(x) = \begin {cases} x^2 &\text {if $x<3$,}\\ mx+b &\text {if $x\ge 3$.} \end {cases}$ What values of $m$ and $b$ make $f$ differentiable at $x=3$ ?

To start, we know that $f$ must be continuous at $x=3$ , since it has to be differentiable there. We will start by making $f$ continuous at $x=3$ . Write with me: $\begin{align*} \lim_{x\to 3^-} f(x) &= \answer[given]{9}\\ \lim_{x\to 3^+} f(x) &= \answer[given]{m 3 + b}\\ f(3) &= \answer[given]{m 3 + b} \end{align*}$ So for the function to be continuous, we must have $m\cdot 3 + b =9.$ We also must ensure that the function is differentiable at $x=3$ . In other words, we have to ensure that the following limit exists

$\lim _{x\to 3}\frac {f(x)-f(3)}{x-3}.\\$ In order to compute this limit, we have to compute the two one-sided limits $\lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3}\\$ and $\lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3},\\$ since $f(x)$ changes expression at $x=3$ . Write with me

$\begin{align*} \lim_{x\to 3^{-}}\frac{f(x)-f(3)}{x-3}&= \lim_{x\to 3}\frac{x^2 -9}{x-3}\\ &= \lim_{x\to 3^{-}}\frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}\\ &= \lim_{x\to 3^{-}}\left(x+3\right)\\ &=6, \end{align*}$ and $\begin{align*} \lim_{x\to 3^{+}}\frac{f(x)-f(3)}{x-3}&= \lim_{x\to 3}\frac{mx+b -(3m+b)}{x-3}\\ &= \lim_{x\to 3^{+}}\frac{mx-3m}{x-3}\\ &= \lim_{x\to 3^{+}}\frac{m\cancel{(x-3)}}{\cancel{x-3}}\\ &= \lim_{x\to 3^{+}}m\\ &=m. \end{align*}$ Hence, we must have $m=6.$ Ah! So now the equation that must be satisfied $\begin{align*} 9 &= m\cdot 3 + b,\\ becomes\hspace{0.3in} 9 &= 6\cdot 3 + b.\\ \end{align*}$ Therefore, $b=\answer [given]{-9}$ . Thus setting $m=\answer [given]{6}$ and $b=\answer [given]{-9}$ will give us a function that is differentiable (and hence continuous) at $x=3$ .

Can we tell from its graph whether the function is differentiable or not at a point $a$ ?

What does the graph of a function $f$ possibly look like when $f$ is not differentiable at $a$ ?

Each of the figures A-D depicts a function that is not differentiable at $a=1$ .

The function in figure A is not continuous at $a$ , and, therefore, it is not differentiable there.

In figures $B$ – $D$ the functions are continuous at $a$ , but in each case the limit $\lim _{x\to a} \frac {f(x)-f(a)}{x-a}$ does not exist, for a different reason.

In figure $B$ $\lim _{x\to a^{+}} \frac {f(x)-f(a)}{x-a}\ne \lim _{x\to a^{-}} \frac {f(x)-f(a)}{x-a}.$

In figure $C$ $\lim _{x\to a} \frac {f(x)-f(a)}{x-a}=\infty .$ In figure $D$ the two one-sided limits don’t exist and neither one of them is infinity.

So, if at the point $a$ a function either has a ”jump” in the graph, or a corner, or what looks like a “vertical tangent line”, or if it rapidly oscillates near $a$ , then the function is not differentiable at $a$ .