Removing Discontinuities

Defining or redefining

f

at one point changes only the value at

a

, and the limit at

a

doesn't depend on that value. So the new value can be set equal to the limit. A jump or an infinite discontinuity has no limit to use as the new value.

Example 1

Each function is undefined at

x = a

. Find the value

f(a)

should be given so that

f

is continuous at

a

\begin{aligned} &\text{(a)}\;\; f(x) = \frac{x^3 - 1}{x^2 - 1}, \quad a = 1 \\[10pt] &\text{(b)}\;\; f(x) = \frac{x^2}{1 - \cos x}, \quad a = 0 \end{aligned}

(a) Factor the difference of cubes and the difference of squares. For

x \ne \pm 1

\begin{aligned} &\frac{x^3 - 1}{x^2 - 1} \\[4pt] &\qquad = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} \\[4pt] &\qquad = \frac{x^2 + x + 1}{x + 1}. \end{aligned}

\begin{aligned} \lim_{x \to 1} f(x) &= \lim_{x \to 1} \frac{x^2 + x + 1}{x + 1} \\[4pt] &= \frac{1 + 1 + 1}{1 + 1} = \frac{3}{2}. \end{aligned}

Defining

f(1) = \frac{3}{2}

makes

f(1) = \lim_{x \to 1} f(x)

, so

f

is continuous at

1

. The other zero of the denominator,

x = -1

, is different: there

x^3 - 1

approaches

-2

while the denominator approaches

0

, so the discontinuity is infinite and no value removes it.

The graph of f: a hole at (1, 3/2), which f(1) = 3/2 fills, and the vertical asymptote x = −1.

(b) Multiply above and below by

1 + \cos x

, and use

1 - \cos^2 x = \sin^2 x

\begin{aligned} \frac{x^2}{1 - \cos x} &= \frac{x^2(1 + \cos x)}{\sin^2 x} \\[4pt] &= \left(\frac{x}{\sin x}\right)^2 (1 + \cos x). \end{aligned}

x

approaches

0

\frac{x}{\sin x}

approaches

1

and

1 + \cos x

approaches

2

, so

\lim_{x \to 0} f(x) = 1^2 \cdot 2 = 2

. Defining

f(0) = 2

makes

f(0) = \lim_{x \to 0} f(x)

, so

f

is continuous at

0

Constants in a piecewise function

Example 3

Find the values of

a

and

b

that make

f

continuous for all

x

f(x) = \begin{cases} ax + 3, & x < 1 \\ x^2 + b, & 1 \le x \le 3 \\ 2ax - 4, & x > 3 \end{cases}

Each piece is a polynomial, so

f

is continuous everywhere except possibly at the seams

x = 1

and

x = 3

. Each seam gives one equation:

\begin{aligned} x = 1:&\quad a + 3 = 1 + b \\[4pt] x = 3:&\quad 9 + b = 6a - 4 \end{aligned}

The first gives

b = a + 2

. Substituting into the second,

11 + a = 6a - 4

, so

a = 3

and

b = 5

. The pieces then meet at

(1, 6)

and at

(3, 14)

With a = 3 and b = 5, the three pieces join at (1, 6) and (3, 14).

A calculator example

Example 5

A graphing calculator is allowed. Find every value of

k

that makes

f

continuous at

x = 1

f(x) = \begin{cases} 2^k x, & x \le 1 \\ k^2 x^3, & x > 1 \end{cases}

Both pieces are continuous, so

f

is continuous at

1

exactly when

2^k = k^2

. Trying small integers finds

k = 2

and

k = 4

, since

2^2 = 4

and

2^4 = 16 = 4^2

. The graph of

y = 2^k - k^2

shows a third zero, for a negative

k

, where

2^k

is between

0

and

1

while

k^2

grows as

k

decreases.

y = 2ᵏ − k² crosses zero three times: near −0.767, and at 2 and 4.

The calculator's zero finder gives

k \approx -0.767

. The graph shows only these three zeros between

-1

and

5

, and there are no others. When

k < -1

, we have

2^k < 1 < k^2

. Past

5

, each step of

1

k

doubles

2^k

but multiplies

k^2

by less than

2

, so

2^k

stays ahead. So

f

is continuous at

1

for

k = 2

k = 4

, and

k \approx -0.767