Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.

1. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Sec on 1.5
Con nuity
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
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February 7, 2011
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Notes
Announcements
Get-to-know-you extra
credit due Friday
February 11
Quiz 1 February 17/18 in
recita on
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Notes
Objectives
Understand and apply the defini on of
con nuity for a func on at a point or
on an interval.
Given a piecewise defined func on,
decide where it is con nuous or
discon nuous.
State and understand the Intermediate
Value Theorem.
Use the IVT to show that a func on
takes a certain value, or that an
equa on has a solu on
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. 1
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2. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Last time
Defini on
We write
lim f(x) = L
x→a
and say
“the limit of f(x), as x approaches a, equals L”
if we can make the values of f(x) arbitrarily close to L (as close to L
as we like) by taking x to be sufficiently close to a (on either side of
a) but not equal to a.
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Notes
Basic Limits
Theorem (Basic Limits)
lim x = a
x→a
lim c = c
x→a
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Notes
Limit Laws for arithmetic
Theorem (Limit Laws)
Let f and g be func ons with limits at a point a. Then
lim (f(x) + g(x)) = lim f(x) + lim g(x)
x→a x→a x→a
lim (f(x) − g(x)) = lim f(x) − lim g(x)
x→a x→a x→a
lim (f(x) · g(x)) = lim f(x) · lim g(x)
x→a x→a x→a
f(x) limx→a f(x)
lim = if lim g(x) ̸= 0
x→a g(x) limx→a g(x) x→a
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3. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Hatsumon
Here are some discussion ques ons to start.
True or False
At some point in your life you were exactly three feet tall.
True or False
At some point in your life your height (in inches) was equal to your
weight (in pounds).
True or False
Right now there are a pair of points on opposite sides of the world
measuring the exact same temperature.
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Notes
Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons
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Recall: Direct Substitution Notes
Property
Theorem (The Direct Subs tu on Property)
If f is a polynomial or a ra onal func on and a is in the domain of f,
then
lim f(x) = f(a)
x→a
This property is so useful it’s worth naming.
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4. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Definition of Continuity
Defini on y
Let f be a func on defined near
a. We say that f is con nuous at f(a)
a if
lim f(x) = f(a).
x→a
A func on f is con nuous if it is
con nuous at every point in its
domain. . x
a
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Notes
Scholium
Defini on
Let f be a func on defined near a. We say that f is con nuous at a if
lim f(x) = f(a).
x→a
There are three important parts to this defini on.
The func on has to have a limit at a,
the func on has to have a value at a,
and these values have to agree.
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Notes
Free Theorems
Theorem
(a) Any polynomial is con nuous everywhere; that is, it is
con nuous on R = (−∞, ∞).
(b) Any ra onal func on is con nuous wherever it is defined; that is,
it is con nuous on its domain.
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5. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Showing a function is continuous
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Example
√
Let f(x) = 4x + 1. Show that f is con nuous at 2.
Solu on
We want to show that lim f(x) = f(2). We have
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2
Each step comes from the limit laws.
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Notes
At which other points?
Ques on
√
As before, let f(x) = 4x + 1. At which points is f con nuous?
Solu on
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Notes
Limit Laws give Continuity Laws
Theorem
If f(x) and g(x) are con nuous at a and c is a constant, then the
following func ons are also con nuous at a:
(f + g)(x) (fg)(x)
(f − g)(x) f
(x) (if g(a) ̸= 0)
(cf)(x) g
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. 5
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6. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Sum of continuous is continuous
We want to show that
lim (f + g)(x) = (f + g)(a).
x→a
We just follow our nose.
(def of f + g) lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
(if these limits exist) = lim f(x) + lim g(x)
x→a x→a
(they do; f and g are cts.) = f(a) + g(a)
(def of f + g again) = (f + g)(a)
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Notes
Trig functions are continuous
tan sec
sin and cos are con nuous
on R.
sin 1
tan = and sec = are
cos cos cos
con nuous on their domain,
{π
which is } .
sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
con nuous on their domain,
which is R { kπ | k ∈ Z }. cot csc
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Notes
Exp and Log are continuous
For any base a 1,
the func on x → ax is ax
loga x
con nuous on R
the func on loga is
con nuous on its .
domain: (0, ∞)
In par cular ex and
ln = loge are con nuous
on their domains
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. 6
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7. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Inverse trigonometric functions Notes
are mostly continuous
sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,
and right con nuous at −1.
sec−1 and csc−1 are con nuous on (−∞, −1) ∪ (1, ∞), le
con nuous at −1, and right con nuous at 1.
tan−1 and cot−1 are con nuous on R.
π
cot−1 cos−1 sec−1
π/2 tan−1
csc−1
sin−1 .
. −π/2
−π .
Notes
What could go wrong?
In what ways could a func on f fail to be con nuous at a point a?
Look again at the equa on from the defini on:
lim f(x) = f(a)
x→a
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Notes
Continuity FAIL: no limit
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Example
{
x2 if 0 ≤ x ≤ 1
Let f(x) = . At which points is f con nuous?
2x if 1 x ≤ 2
Solu on
At any point a besides 1, lim f(x) = f(a) because f is represented by a
x→a
polynomial near a, and polynomials have the direct subs tu on property.
lim f(x) = lim− x2 = 12 = 1 and lim+ f(x) = lim+ 2x = 2(1) = 2
x→1− x→1 x→1 x→1
So f has no limit at 1. Therefore f is not con nuous at 1.
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. 7
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8. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Graphical Illustration of Pitfall #1
y
4
3 The func on cannot be
con nuous at a point if the
2
func on has no limit at that
1 point.
. x
−1 1 2
−1
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Notes
Continuity FAIL: no value
Example
Let
x2 + 2x + 1
f(x) =
x+1
At which points is f con nuous?
Solu on
Because f is ra onal, it is con nuous on its whole domain. Note that
−1 is not in the domain of f, so f is not con nuous there.
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Notes
Graphical Illustration of Pitfall #2
y
1 The func on cannot be
con nuous at a point outside
. x its domain (that is, a point
−1 where it has no value).
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9. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Continuity FAIL: value ̸= limit
Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1
At which points is f con nuous?
Solu on
f is not con nuous at 1 because f(1) = π but lim f(x) = 7.
x→1
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Notes
Graphical Illustration of Pitfall #3
y
7 If the func on has a limit and
a value at a point the two
π must s ll agree.
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1
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Notes
Special types of discontinuities
removable discon nuity The limit lim f(x) exists, but f is not
x→a
defined at a or its value at a is not equal to the limit at a.
By re-defining f(a) = lim f(x), f can be made con nuous
x→a
at a
jump discon nuity The limits lim− f(x) and lim+ f(x) exist, but are
x→a x→a
different. The func on cannot be made con nuous by
changing a single value.
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10. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Special discontinuities graphically
y y
7 2
π 1
. x . x
1 1
removable jump
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Notes
The greatest integer function
[[x]] is the greatest integer ≤ x. y
3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1 . x
2.1 2 −2 −1 1 2 3
−0.5 −1 −1
−0.9 −1
−1.1 −2 −2
This func on has a jump discon nuity at each integer.
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Notes
Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons
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11. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is con nuous on the closed interval [a, b] and let N be
any number between f(a) and f(b), where f(a) ̸= f(b). Then there
exists a number c in (a, b) such that f(c) = N.
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Notes
Illustrating the IVT
f(x)
Theorem
Suppose that f is con nuous f(b)
on the closed interval [a, b]
N
and let N be any number
between f(a) and f(b), where f(a)
f(a) ̸= f(b). Then there exists
a number c in (a, b) such that
f(c) = N. . a c b x
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Notes
What the IVT does not say
The Intermediate Value Theorem is an “existence” theorem.
It does not say how many such c exist.
It also does not say how to find c.
S ll, it can be used in itera on or in conjunc on with other
theorems to answer these ques ons.
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12. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Using the IVT to find zeroes
Example
Let f(x) = x3 − x − 1. Show that there is a zero for f on the interval
[1, 2].
Solu on
f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.
In fact, we can “narrow in” on the zero by the method of bisec ons.
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Notes
Finding a zero by bisection
y
x f(x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5
. x
(More careful analysis yields
1.32472.)
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Using the IVT to assert existence Notes
of numbers
Example
Suppose we are unaware of the square root func on and that it’s
con nuous. Prove that the square root of two exists.
Proof.
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13. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Outline
Con nuity
The Intermediate Value Theorem
Back to the Ques ons
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Notes
Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
True or False
At one point in your life your height in inches equaled your weight in
pounds.
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.
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Notes
Question 1
To be discussed in class!
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14. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Notes
Question 2
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Notes
Question 3
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Notes
Question 3
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15. . V63.0121.001: Calculus I
. Sec on 1.5:. Limits February 7, 2011
Summary Notes
What have we learned today?
Defini on: a func on is con nuous at a point if the limit of the
func on at that point agrees with the value of the func on at
that point.
We o en make a fundamental assump on that func ons we
meet in nature are con nuous.
The Intermediate Value Theorem is a basic property of real
numbers that we need and use a lot.
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Notes
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Notes
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