1.4 Continuity and One-Sided Limits
Definition of Continuity:
Informal: A function is continuous if its graph can be drawn without lifting the pencil from the paper.
Formal Definition of Continuity: A function is continuous at c if:
1) f(c) is defined
2) lim f ( x) exists
x!c
3) lim f ( x) = f (c)
x!c
1.5
1
0.5
c
-1
1
-0.5
Types of Discontinuities:
Removable Discontinuity: If f can be made continuous by defining or redefining f(c) then you have a
removable discontinuity (i.e. a point discontinuity).
( x + 2)( x + 3)
Examine the graph of this function and redefine it as a piecewise
( x + 2)
function that is continuous.
Example 1: f ( x) =
Nonremovable Discontinuity: The function f can not be defined in any way to make it continuous (i.e.
you have an asymptote or large break in the graph).
Example 2: f ( x) =
1
Examine the graph of this function and observe the discontinuity.
x!5
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One Sided Limits:
lim f ( x) = L
limit as x approaches c from the right is L
lim f ( x) = L
limit as x approaches c from the left is L
x!c +
x"c !
Determine the following limits:
x
Example 3: lim+ =
x!0 x
lim!
x"0
x
=
x
Example 4: lim+ x 2 + 3 =
x !2
lim x 2 + 3 =
x "2!
Existence of a Limit
Let f be a function and let c and L be real numbers. The limit of f(x) as x approaches c is L if and only if,
lim f ( x) = L and lim+ f ( x) = L
x"c !
(see example 4 above)
x!c
Intermediate Value Theorem
If f is continuous on the closed interval [a,b] and k is any number between f(a) and f(b), then there is at least one
number c in [a,b] such that f(c) = k.
25
f(a)
20
15
k
f(a)
k
f(b)
15
10
10
f(b)
5
5
a c
-1
1
2
b
3
a c1
-1
1
2
c2
-5
-5
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3
4
b
5
Homework Examples:
Find the limit, if it exists:
1.
4.
lim [x ]+ 2
lim! 9 ! x 2
2.
lim[x ]+ 2
# x &
5. lim+ % 2
(
x! 7 $ x " 49 '
x"3
x!3
lim [x ]+ 2
3.
x"3!
x!3+
6. lim" 2x + 3
x!2
Find the discontinuities, if they are present. Are they removable or non-removable discontinuities? If there are
removable discontinuities, rewrite the function as a piecewise function that is continuous.
1
x +2
5.
f ( x) =
6.
f ( x) = tan x
7.
f ( x) =
4
x 2 ! 7 x ! 30
x+3
Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of c
guaranteed by the theorem.
1. f ( x) = x 2 ! 6 x + 8 [0,3] f(c) = 0
2. f (x) = x 3 ! 3x 2 + 2x ! 1 [3, 5]
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f(c)= 23