MATH 136 Continuity: Limits of Piecewise-Defined Functions

Given a piecewise-defined function that is “split” at some point x = a , we wish to determine if lim

x→af ( x) exists and to determine if f is continuous at x = a .

Recall: In order for lim

x→af ( x) to exist, both lim

x→a−f (x ) and lim

x→a+f (x ) must exist as

finite numbers and they must be equal. If these one-sided limits both equal L , then

€

limx→ a

f (x) = L also. The actual function value f (a) is irrelevant with regard to

evaluating the limit. If either one-sided limit does not exist or if the one-sided limits are not equal, then limx→a

f ( x) does not exist.

Checking Continuity

Definition. Let f be a function and let a be a point in its domain. Then f is continuous at the single point x = a provided

limx→a

f ( x) = f (a) .

If f is continuous at each point in its domain, then we say that f is continuous.

Many functions are continuous such as sin x , cos x , ex , ln x , and any polynomial. Other functions are continuous over certain intervals such as tan x for − π

2< x <

π

2. For

a continuous function, we evaluate limits easily by direct substitution. For example, limx→3

x2 = 32 = 9 .

But we are concerned now with determining continuity at the point x = a for a

piecewise-defined function of the form f ( x) =

f1(x ) if x < a c if x = af 2 (x ) if x > a

.

For a function of this form to be continuous at x = a , we must have: (i) lim

x→a−f (x ) and lim

x→a+f (x ) must exist and be equal (that is, lim

x→af ( x) must exist);

(ii) f (a) must be defined; and (iii) f (a) must equal lim

x→af ( x) .

Types of Discontinuities

If a piecewise-defined function f is not continuous at x = a , then there is a discontinuity which can take one of the following forms: (i) If lim

x→af ( x) exists, but f (a) is either not defined or does not equal the limit. Then

there is a “hole in the graph,” which is formally called a removable discontinuity. (ii) If the one-sided limits are finite but not equal, lim

x→a−f (x ) ≠ lim

x→a+f (x ) , then there is

a jump discontinuity, which is also called a non-removable discontinuity. (iii) If one or both of the one-sided limits is infinite, then there is a vertical asymptote, which is called an infinite discontinuity.

Example 1. Let f ( x) =

3−x − 20 if x < −3 5 if x = −37 sin(π x / 2) if x > −3

.

(a) Evaluate the limits:

(i) limx→−3−

f (x ) (ii) limx→−3+

f (x ) (iii) limx→−3

f (x )

(b) Explain whether or not f is continuous at x = −3 . If f is not continuous at this point, then explain what kind of discontinuity there is.

y = 3−x − 20

y = 7sin(π x / 2)

•

f ( x) =

3−x − 20 if x < −3 5 if x = −37 sin(π x / 2) if x > −3

Solution. From the left, we have lim

x→−3−f (x ) = lim

x→−3−(3− x − 20) = 33 − 20 = 7 . From the

right, we have limx→−3+

f (x ) = limx→−3−

7sin(π x / 2) = 7sin(−3π / 2) = 7. Because these one-

sided limits are finite and equal, limx→−3

f (x ) exists and limx→−3

f (x ) = 7 also. But not that

f (−3) = 5 ≠ limx→−3

f ( x) . Thus, f is not continuous at x = −3 .

Because limx→−3

f (x ) exists but does not equal f (−3) , there is a “hole” in the graph

which is a removable discontinuity. This type of discontinuity is called removable because we could re-define f (−3) as f (−3) = 7 in order to fill the hole and remove the discontinuity.

Example 2. Let f ( x) =

4ln(x − 3) if x > 4 3 if x = 43cos(x − 4) if x < 4

.

(a) Evaluate the limits:

(i) lim

x→ 4−f ( x) (ii) lim

x→ 4+f ( x) (iii) lim

x→ 4f ( x)

(b) Explain whether or not f is continuous at x = 4 . If f is not continuous at this point, then explain what kind of discontinuity there is. Solution. From the left we have lim

x→ 4−f ( x) = lim

x→ 4−3cos(x − 4) = 3cos 0 = 3 , and from

the right limx→ 4+

f ( x) = limx→ 4+

4ln( x − 3) = 4 ln1 = 0 . Because limx→ 4−

f ( x) ≠ limx→ 4+

f ( x) ,

limx→ 4

f ( x) does not exist.

Because limx→ 4

f ( x) does not exist, f is not continuous at x = 4 . Because the one-

sided limits are different, there is a jump (non-removable) discontinuity. Note: Because lim

x→ 4−f ( x) = f (4) ,

we can say that f is left-continuous at x = 4 . 3cos(x − 4) 4 ln( x − 3)

•

Example 3. Let f ( x) =

x1/3 if x < 8 2 if x = 8x − 4 if x > 8

.

(a) Evaluate the limits:

(i) limx→ 8−

f (x ) (ii) limx→ 8+

f (x ) (iii) limx→ 8

f ( x)

(b) Explain whether or not f is continuous at x = 8. If f is not continuous at this point, then explain what kind of discontinuity there is. Solution. From the left, lim

x→ 8−f (x ) = lim

x→ 8−x1/3 = 81/3 = 2 and from the right

limx→ 8+

f (x ) = limx→ 8+

x − 4 = 4 = 2 . Because these one-sided limits are finite and equal,

limx→ 8

f ( x) exists and limx→ 8

f ( x) = 2 also. Moreover, f (8) = 2 . So because

limx→ 8

f ( x) = f (8) , f is continuous at x = 8.

•

Creating a Discontinuity in a Constant Function

Given a constant function f ( x) = c , we can create a “hole in the graph” at x = a by

multiplying by the term (x − a) / ( x − a) . The resulting function ˜ f ( x) = c( x − a)( x − a)

is not

defined at x = a although limx→ a

˜ f ( x) = c . There is a removable discontinuity at x = a

and ˜ f ( x) = c for x ≠ a .

If instead we multiply by x − a / ( x − a) , then we split the constant into a step function that has a non-removable discontinuity at x = a . Because x − a / ( x − a) = 1 for x > a and x − a / ( x − a) = –1 for x < a , we have

ˆ f ( x) =c x − a(x − a)

= c if x > a

−c if x < a .

Continuity of Rational Functions

A rational function has the form f ( x) = p( x)q(x )

, where p(x ) and q(x ) are polynomials.

Then f ( x) is not defined whenever q(x ) = 0 , so f cannot be continuous at these points. These discontinuities will either be asymptotes or removable. If q(a) = 0 but p(a) ≠ 0 , then f will have a vertical asymptote at x = a . However if q(a) = 0 and p(a) = 0 , then there will be a removable discontinuity at x = a provided the multiplicity of this root for p(x ) is greater than or equal to the multiplicity of this root for q(x ) . But if the multiplicity of the root for p(x ) is less than the multiplicity of the root for q(x ) , then there will be vertical asymptote at x = a .

Example 4. Let f ( x) = x2 + x − 6x2 − x − 12

. Determine and describe all discontinuities.

Solution. We can factor f as f ( x) = ( x − 2)( x + 3)

( x − 4)( x + 3)=

(x − 2)(x − 4)

for x ≠ −3 . The roots of the

denominator are x = 4 and x = −3 , so f is discontinuous at these points. However x = −3 is also a root of the numerator with equal multiplicity, so all terms involving x − (−3) cancel out of both the numerator and denominator. Thus,

limx→−3

f (x ) = limx→−3

( x − 2)( x − 4)

=57

,

and f has a removable discontinuity at x = −3 . On the other hand, x = 4 is not a root of the numerator, so there will be a vertical asymptote (infinite discontinuity) at x = 4 . We now have

limx→ 4−

f ( x) = limx→ 4−

(x − 2)(x − 4)

=2−0

= −∞ and limx→ 4+

f ( x) = limx→ 4+

(x − 2)(x − 4)

=2+0

= +∞ .

Example 5. Let g(x ) = x 2 −10 x +16x4 − 8x2 + 16

. Determine and describe all discontinuities.

Solution. We factor g as

g(x ) = x 2 −10 x +16x4 − 8x2 + 16

=( x − 2)(x − 8)( x2 − 4)(x2 − 4)

=(x − 2)(x − 8)( x − 2)2 (x + 2)2

.

The roots of the denominator are x = 2 and x = −2 , both having multiplicity 2. Because x = −2 is not a root of the numerator, we have a vertical asymptote (infinite discontinuity) at x = −2 . We then have

limx→−2−

g( x) = limx→−2−

(x − 8)(x − 2)(x + 2)2

=−10−0

= +∞

and

lim

x→−2+g( x) = lim

x→−2+

(x − 8)(x − 2)(x + 2)2

=−10−0

= +∞ .

On the other hand, x = 2 is a root of the numerator, but only with multiplicity 1. So the term x − 2 does not completely cancel out of the denominator which causes another vertical asymptote at x = 2 . We now have

lim

x→ 2−g( x) = lim

x→2−

( x − 8)( x − 2)( x + 2)2

=−6−0

= +∞

and

lim

x→ 2+g( x) = lim

x→2+

( x − 8)( x − 2)( x + 2)2

=−6+0

= −∞ .