Continuity and Differentiability

Without lifting the pen

Your first idea of continuity is probably the right one: a function is continuous at a point if you can draw its graph through that point without lifting your pen. Here we make that idea precise using limits. You will test piecewise functions, find the constants that make a function continuous, and see how continuity is related to differentiability.

Continuity

$f$ is continuous at $c$ (in its domain) if

$\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c).$

We call a function continuous if it is continuous at every point of its domain. Polynomials, $\sin x$ , $\cos x$ , $e^x$ and $\lvert x \rvert$ are continuous everywhere, and rational functions are continuous wherever the denominator is non-zero ( $\displaystyle \tfrac{1}{x}$ is continuous on its domain, which excludes $0$ ).

This saves a lot of work: add, subtract, multiply or divide continuous functions (with a non-zero denominator) and the result is continuous. The same goes for composites.

Example 1. $f(x) = \begin{cases} 3x - 1, & x \le 2 \\ x^2 + 1, & x > 2 \end{cases}$ . The left limit is $5$ , the right limit is $5$ , and $f(2) = 5$ . So it is continuous at $2$ , and hence everywhere.

Graph of the line y = 3x - 1 up to x = 2 joined to the curve y = x squared + 1 after x = 2; both pieces reach the point (2, 5), so the graph has no break. — Both pieces reach (2, 5): the graph is drawn without lifting the pen.

Example 2. $\displaystyle f(x) = \begin{cases} \frac{x^2 - 16}{x - 4}, & x \ne 4 \\ 7, & x = 4 \end{cases}$ . Here $\displaystyle \lim_{x \to 4} f(x) = 8 \ne 7$ , so the function is discontinuous at $4$ .

The line y = x + 4 with an open circle at (4, 8), where the limit is 8, and a separate filled dot at (4, 7) showing the value f(4) = 7 below the gap. — The limit at 4 is 8, but the value there is 7.

Example 3. The greatest integer function $[x]$ is discontinuous at every integer. At $n$ , the left limit is $n - 1$ but the right limit is $n$ .

Step graph of y = [x] from -3 to 3: flat steps with a filled dot at each left end and an open circle at each right end; at x = 2 the left limit is 1 and the right limit 2. — The greatest integer function jumps up by 1 at every integer.

Example 4 (finding a constant). $f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$ is continuous at $\pi$ when the two pieces meet, that is, when $k\pi + 1 = -1$ , which gives $\displaystyle k = -\tfrac{2}{\pi}$ .

Example 5. $f(x) = \begin{cases} ax + 1, & x \le 1 \\ 5, & 1 < x < 3 \\ bx - 4, & x \ge 3 \end{cases}$ . Matching the pieces at $1$ gives $a + 1 = 5$ , and at $3$ gives $3b - 4 = 5$ . So $a = 4$ , $b = 3$ .

Differentiability

$f$ is differentiable at $c$ if $\displaystyle f'(c) = \lim_{h \to 0}\tfrac{f(c + h) - f(c)}{h}$ exists, which means the left and right difference quotients agree.

Every differentiable function is continuous, but the reverse is false. Take $\lvert x \rvert$ : it is continuous at $0$ , yet the left and right quotients are $-1$ and $1$ , so it has no derivative there. You can see the reason on the graph, a sharp corner.

Graph of y = |x|, a V shape with slope -1 on the left and slope 1 on the right, meeting in a sharp corner at the origin where there is no derivative. — y = |x| is continuous at 0 but has a corner there.

Example 6. $f(x) = \lvert x - 3 \rvert$ is continuous everywhere but not differentiable at $3$ . On the other hand, $f(x) = x\lvert x \rvert$ is differentiable at $0$ , since both quotients tend to $0$ .

Try these yourself

Is $f(x) = 2x^2 - 5x + 3$ continuous at $x = 1$ ? Everywhere?
Examine continuity of $f(x) = \begin{cases} x + 4, & x < 1 \\ 5x, & x \ge 1 \end{cases}$ at $x = 1$ .
Find all points of discontinuity of $\displaystyle f(x) = \frac{x + 2}{x^2 - 9}$ and of $[2x]$ on $[0, 2]$ .
Find $k$ so that $\displaystyle f(x) = \begin{cases} \frac{\sin 3x}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $0$ .
Find $a$ and $b$ so that $f(x) = \begin{cases} 2, & x \le 0 \\ ax + b, & 0 < x < 4 \\ 10, & x \ge 4 \end{cases}$ is continuous everywhere.
Is $\displaystyle f(x) = \begin{cases} x^2 \sin\frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ continuous at $0$ ?
Show that $f(x) = \lvert x + 2 \rvert$ is continuous but not differentiable at $x = -2$ .
Is $f(x) = \lvert x \rvert + \lvert x - 1 \rvert$ differentiable at $0$ and $1$ ?
Give a function that is continuous on $\mathbf{R}$ but not differentiable at $x = 5$ .
Is $f(x) = [x]$ differentiable at $x = 2.5$ ? At $x = 2$ ?

Answers to check against

Show answers

Yes; yes (polynomial).
The left limit is $5$ , the right $5$ , and $f(1) = 5$ , so it is continuous.
$x = \pm 3$ ; $\displaystyle x = \tfrac{1}{2}, 1, \tfrac{3}{2}, 2$ (and $0$ from the right is fine).
$k = 3$ .
$b = 2$ , $4a + 2 = 10$ : $a = 2$ .
Yes, because $\displaystyle \lvert x^2\sin\tfrac{1}{x} \rvert \le x^2 \to 0$ .
Left and right limits are $0 = f(-2)$ ; the one-sided quotients are $-1$ and $1$ .
Not differentiable at either (both are corners).
$\lvert x - 5 \rvert$ .
Yes (derivative $0$ ); no, since it is not even continuous there.

Menu

Without lifting the pen

Continuity

Differentiability

Try these yourself

Answers to check against

Test yourself

Menu

With­out lift­ing the pen#

Con­ti­nu­ity#

Dif­fer­en­tia­bil­ity#

Try these your­self#

Answers to check against#

Test yourself

Without lifting the pen

Continuity

Differentiability

Try these yourself

Answers to check against