MATH 136
Continuity: Limits of PiecewiseDefined Functions
Given a piecewise-defined function that is “split” at some point x = a , we wish to
determine if lim f ( x) exists and to determine if f is continuous at x = a .
x →a
Recall: In order for lim f ( x) to exist, both
x →a
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
lim f (x) = L also. The actual function value f (a) is irrelevant with regard to
x→a
evaluating the limit.
€
If either one-sided limit does not exist or if the one-sided limits are not equal, then
lim f ( x) does not exist.
x →a
Checking Continuity
Definition. Let f be a function and let a be a point in its domain.
continuous at the single point x = a provided
Then f is
lim f ( x) = f (a) .
x →a
If f is continuous at each point in its domain, then we say that f is continuous.
x
Many functions are continuous such as sin x , cos x , e , ln x , and any polynomial.
π
π
Other functions are continuous over certain intervals such as tan x for − < x < . For
2
2
a continuous function, we evaluate limits easily by direct substitution. For example,
2
2
lim x = 3 = 9 .
x →3
But we are concerned now with determining continuity at the point x = a for a
 f 1 (x ) if x < a

piecewise-defined function of the form f ( x) =  c
if x = a .

 f 2 (x ) if x > a
For a function of this form to be continuous at x = a , we must have:
(i) lim f (x ) and lim f (x ) must exist and be equal (that is, lim f ( x) must exist);
x →a−
x →a+
(ii) f (a) must be defined; and
(iii) f (a) must equal lim f ( x) .
x →a
x →a
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 f ( x) exists, but f (a) is either not defined or does not equal the limit. Then
x →a
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 f (x ) ≠ lim f (x ) , then there is
x →a−
x →a+
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.
 3 −x − 20

5
Example 1. Let f ( x) = 

7 sin(π x / 2)
if x < −3
if x = −3 .
if x > −3
(a) Evaluate the limits:
(i)
lim
x → −3−
f (x )
(ii)
lim
x → −3+
f (x )
(iii) lim f (x )
x → −3
(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 = 7sin(π x / 2)
y = 3−x − 20
 3 −x − 20

f ( x) = 
5

7 sin(π x / 2)
•
if x < −3
if x = −3
if x > −3
Solution. From the left, we have
right, we have
lim
x → −3+
f (x ) =
lim
x → −3−
lim
x → −3 −
f (x ) =
lim
x → −3−
(3
−x
3
− 20) = 3 − 20 = 7 . From the
7sin(π x / 2) = 7sin(−3π / 2) = 7 . Because these one-
sided limits are finite and equal, lim f (x ) exists and lim f (x ) = 7 also. But not that
x → −3
x → −3
f (−3) = 5 ≠ lim f ( x) . Thus, f is not continuous at x = −3 .
x → −3
Because
lim f (x ) exists but does not equal f (−3) , there is a “hole” in the graph
x → −3
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.
 4ln(x − 3)

Example 2. Let f ( x) = 
3

3cos(x − 4)
if x > 4
if x = 4 .
if x < 4
(a) Evaluate the limits:
(i)
lim
x → 4−
(ii) lim
f ( x)
x → 4+
f ( x)
(iii) lim f ( x)
x→ 4
(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
the right
lim
x → 4+
lim
x → 4−
f ( x) = lim 3cos(x − 4) = 3cos 0 = 3 , and from
x → 4−
f ( x) = lim 4ln( x − 3) = 4 ln1 = 0 . Because
x → 4+
lim
x → 4−
f ( x) ≠
lim
x → 4+
f ( x) ,
lim f ( x) does not exist.
x→ 4
Because lim f ( x) does not exist, f is not continuous at x = 4 . Because the onex→ 4
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)
 x 1/3

2
Example 3. Let f ( x) = 

 x − 4
if x < 8
if x = 8 .
if x > 8
(a) Evaluate the limits:
(i)
(ii) lim f (x )
lim f (x )
x → 8−
x → 8+
(iii) lim f ( x)
x→ 8
(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 f (x ) = lim
x → 8+
x → 8+
lim
x → 8−
f (x ) = lim x 1/3 = 81/ 3 = 2 and from the right
x → 8−
x − 4 = 4 = 2 . Because these one-sided limits are finite and equal,
lim f ( x) exists and
x→ 8
lim f ( x) = 2 also.
x→ 8
Moreover,
f (8) = 2 .
So because
lim f ( x) = f (8) , f is continuous at x = 8 .
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
c( x − a)
multiplying by the term (x − a) / ( x − a) . The resulting function f˜ ( x) =
is not
( x − a)
defined at x = a although lim f˜ ( x) = c . There is a removable discontinuity at x = a
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
c x − a  c if x > a
fˆ ( x) =
=
(x − a) − c if x < a.
Continuity of Rational Functions
p( x)
, where p(x ) and q(x ) are polynomials.
q(x )
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 .
A rational function has the form f ( x) =
x2 + x − 6
Example 4. Let f ( x) = 2
. Determine and describe all discontinuities.
x − x − 12
( x − 2)( x + 3) (x − 2)
=
for x ≠ −3 . The roots of the
( x − 4)( x + 3) (x − 4)
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,
Solution. We can factor f as f ( x) =
( x − 2) 5
= ,
7
x → −3 ( x − 4)
lim f (x ) = lim
x → −3
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
lim
x → 4−
(x − 2) 2
(x − 2) 2
=
= −∞ and
lim f ( x) = lim
=
= +∞ .
x → 4 − (x − 4) −0
x → 4+
x → 4 + (x − 4) +0
f ( x) = lim
x 2 − 10 x + 16
g(x
)
=
Example 5. Let
. Determine and describe all discontinuities.
x 4 − 8x 2 + 16
Solution. We factor g as
x 2 − 10 x + 16
( x − 2)(x − 8)
(x − 2)(x − 8)
g(x ) = 4
= 2
=
2
2
2
2 .
x − 8x + 16 ( x − 4)(x − 4) ( x − 2) (x + 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
lim
g( x) =
x → −2 −
(x − 8)
−10
=
= +∞
2
−0
x → −2 − (x − 2)(x + 2)
lim
and
lim
x → −2 +
g( x) =
(x − 8)
−10
=
= +∞.
2
−0
x → −2 + (x − 2)(x + 2)
lim
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
( x − 8)
−6
=
= +∞
2
−0
x → 2 − ( x − 2)( x + 2)
lim g( x) = lim
x → 2−
and
( x − 8)
−6
=
= −∞ .
2
+0
x → 2 + ( x − 2)( x + 2)
lim g( x) = lim
x → 2+