MATH 136 Continuity: Limits of Piecewise- Defined …people.wku.edu/david.neal/136/Unit1/Continuity.pdf · MATH 136 Continuity: Limits of Piecewise-Defined Functions Given a piecewise-defined
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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
= −∞ .
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