1.1 An introduction to limits
Definition
(Informal definition)
If we can make the values of a function f (x) get arbitrarily close to a
number L by taking x-values that are sufficiently close to c (but not
equal to c), then we say the limit of f (x) as x approaches c is L, and we
write
lim f (x) = L
x→c
Remark
We need to define more clearly what is means to be arbitrarily close
and sufficiently close in order to make the informal definition of limit
given above more rigorous. We’ll do that in the next section. For now,
we’ll use this informal definition to explore the idea of a limit by
estimating limits (a) graphically and (b) numerically.
1.1 An introduction to limits
Estimating limits graphically
Example
Find
• lim
√
3x + 1 =
• lim
√
3x + 1 =
• lim
√
3x + 1 =
• lim
√
3x + 1 =
• lim
√
3x + 1 =
x→5
x→1
x→0
x→8
x→2
y=
√
3x + 1
1.1 An introduction to limits
y=
√
3x + 1
1.1 An introduction to limits
Estimating limits graphically
Example
Find
• lim
√
3x + 1 = 4
• lim
√
3x + 1 = 2
• lim
√
3x + 1 = 1
• lim
√
3x + 1 = 5
• lim
√
3x + 1 ≈ 2.7
x→5
x→1
x→0
x→8
x→2
y=
√
3x + 1
1.1 An introduction to limits
Estimating limits numerically
Example
2x2 − 2x − 12
x→3
x−3
Find lim
Solution
Pick x-values near 3 (but not equal to 3) and compute the value of the
function.
2x2 −2x−12
2x2 −2x−12
x
x
x−3
x−3
2.9
3.1
2.99
3.01
2.999
3.001
How close to 3 would x have to be in order to get within
• 0.0002 of the limit?
• 0.1 of the limit?
1.1 An introduction to limits
Estimating limits numerically
Example
2x2 − 2x − 12
x→3
x−3
Find lim
Solution
Pick x-values near 3 (but not equal to 3) and compute the value of the
function.
2x2 −2x−12
2x2 −2x−12
x
x
x−3
x−3
2.9
9.8
3.1
10.2
2.99 9.98
3.01 10.02
2.999 9.998
3.001 10.002
How close to 3 would x have to be in order to get within
• 0.0002 of the limit? (within 0.0001 of 3)
• 0.1 of the limit? (within 0.05 of 3)
1.1 An introduction to limits
When limits do not exist
Remark
A function may not have a limit for all values of x. There are three
ways a limit may fail to exist as x approaches c:
1
2
3
The function may approach different values on either side of c or
fail to exist near c.
The function may grow without bound (i.e. tend toward +∞ or
−∞) as x approaches c.
The function may oscillate infinitely often as x approaches c
without staying near any single value.
In the following slides we give examples of each case.
1.1 An introduction to limits
When limits do not exist: (1) different values from left and right
Example
Consider
|x|
=
f (x) =
x
(
x
x =1
−x
x = −1
x>0
x<0
y = |x|/x
|x|
|x|
does not exist because lim+
= 1 (the right-hand limit) while
x→0 x
x→0 x
|x|
= −1 (the left-hand limit).
lim−
x→0 x
lim
1.1 An introduction to limits
When limits do not exist: (1) function fails to exist near c
Example
Consider
f (x) =
√
3x + 1
√
y = 3x + 1
√
√
3x + 1 does not exist because f (x) = 3x + 1 does not exist
lim
x→−1/3
for x < −1/3.
1.1 An introduction to limits
When limits do not exist: (2) function grows without bound
Example
Consider
f (x) =
1
(x − 2)2
y = 1/(x − 2)2
lim 1/(x − 2)2 does not exist because there is no number L that the
x→2
function 1/(x − 2)2 gets closer to as x approaches 2; indeed, the values
of the function 1/(x − 2)2 can be made as large as we’d like by taking
x sufficiently close to 2. In other words, the function grows without
bound as x approaches 2.
1.1 An introduction to limits
When limits do not exist: (3) function oscillates without staying near
a single value
Example
Consider
f (x) = sin(1/x)
y = sin(1/x)
lim sin(1/x) does not exist because there is no number L that the
x→0
function sin(1/x) gets closer to as x approaches 0; indeed, the values
of the function sin(1/x) oscillate more rapidly between −1 and 1 as x
approaches 0.
1.1 An introduction to limits
Example
On the other hand, consider
f (x) = x sin(1/x)
y = x sin(1/x)
lim x sin(1/x) does exist because, even though the function oscillates
x→0
infinitely often near x = 0, the function values are getting closer and
closer to zero.
1.1 An introduction to limits
Limits of difference quotients
Remark
A difference quotient computes the slope of a secant line between two
points (a, f (a)) and (a + h, f (a + h)):
msec =
f (a + h) − f (a)
h
1.1 An introduction to limits
Limits of difference quotients
Remark
Taking the limit of these difference quotients as h → 0 will give us the
slope of the line that is tangent to y = f (x) at the point (a, f (a)):
f (a + h) − f (a)
h→0
h
mtan = lim
Link: GeoGebra tangent line slope.ggb
1.1 An introduction to limits
Just checking. . . .
1
True or false: The limit of f (x) as x approaches 5 is f (5).
2
Approximate lim f (x) numerically and graphically, where
x→0
(
)
cos x
x≤0
f (x) =
, or state why the limit does not
x2 + 3x + 1 x > 0
exist.
x2 + 7x + 10
Approximate lim 2
numerically and graphically, or
x→2 x − 4x + 4
state why the limit does not exist.
3
4
5
Approximate lim f (x) numerically and graphically, where
x→2
(
)
x+1 x<2
f (x) =
, or state why the limit does not exist.
3x − 5 x ≥ 2
Consider the function f (x) = ln x and the point a = 5.
f (a + h) − f (a)
Approximate the limit of the difference quotient
h
by using h = ±0.1, ±0.01.
1.2 Epsilon-delta definition of a limit
Toward a more rigorous definition
Definition
(Informal definitions) Given a function y = f (x), an x-value c, and a
y-value L, we say that lim f (x) = L provided
x→c
1
2
y = f (x) is near L whenever x is near c.
whenever x is within a certain tolerance level of c, then the
corresponding value y = f (x) is within a certain tolerance level of
L.
Remark
The tolerances for x and y are different. The y-tolerance is called ε,
and the x-tolerance is called δ.
1.2 Epsilon-delta definition of a limit
A rigorous definition of limit
Definition
(Rigorous definition) Let f be a function defined on an open interval
containing c (except perhaps at c itself). Then lim f (x) = L provided
x→c
that given any ε > 0, there exists a corresponding δ > 0 such that
whenever 0 < |x − c| < δ, we have |f (x) − L| < ε.
1.2 Epsilon-delta definition of a limit
Examples
Example
1
Show lim
x→9
√
x = 3. (Cf. §1.2: Example 6)
2
Show lim x2 = 16. (Cf. §1.2: Example 7)
3
Show lim ex = 1. (Cf. §1.2: Example 9)
4
Show lim ex = ec .
x→4
x→0
x→c
Remark
This last example show that the function f (x) = ex is continuous at all
values of x. More generally, a function f (x) is continuous at c if
lim f (x) = f (c). We’ll explore this important idea of continuity more
x→c
in §1.5.
1.2 Epsilon-delta definition of a limit
Just checking. . . .
1 What’s wrong with the following “definition” of a limit?
The limit of f (x) as x approaches c is K means that given any
δ > 0, there exists an ε > 0 such that whenever |f (x) − K| < ε we
have 0 < |x − c| < δ.
2
Using an ε − δ argument, show lim 5 = 5.
3
Using an ε − δ argument, show lim 3 − 2x = −1.
x→3
x→2
4
Using an ε − δ argument, show lim x2 − 3 = 6.
x→3
5
Using an ε − δ argument, show lim e2x − 1 = 0.
x→0
1.3 Finding limits analytically
Limit Laws
Theorem
Let b, c, L and K be real numbers, let n be a positive integer, and let f
and g be functions with the following limits:
lim f (x) = L lim g(x) = K
x→c
x→c
Then the following limits hold:
1
2
3
4
5
6
lim b = b
x→c
7
lim x = c
x→c
lim[f (x) ± g(x)] = L ± K
8
x→c
lim b · f (x) = bL
x→c
lim f (x) · g(x) = LK
x→c
lim f (x)/g(x) = L/K(K , 0)
x→c
9
lim[f (x)]n = Ln
x→c
lim
x→c
p
n
f (x) =
√n
L
And if lim f (x) = L and
x→c
lim g(x) = K, then
x→L
lim g(f (x)) = K.
x→c
1.3 Finding limits analytically
Using Limit Laws
Example
Suppose lim f (x) = 3 and lim g(x) = −2 and p(x) = x2 − 3x + 1. Find:
x→2
1
x→2
lim f (x) + g(x) =
3
lim [g(x)]3 =
4
x→2
2
x→2
lim p(x) =
x→2
lim f (x) − 3p(x) =
x→2
1.3 Finding limits analytically
Using Limit Laws
Example
Suppose lim f (x) = 3 and lim g(x) = −2 and p(x) = x2 − 3x + 1. Find:
x→2
1
x→2
lim f (x) + g(x) = 1
3
lim [g(x)]3 = −8
4
x→2
2
lim p(x) = −1
x→2
x→2
lim f (x) − 3p(x) = 6
x→2
Remark
Notice that
lim p(x) = −1 = p(2)
x→2
Whenever lim f (x) = f (c), we say that f is continuous at c. We will
x→c
study continuity in greater detail in §1.5.
1.3 Finding limits analytically
Other Limit Laws
Theorem
Let p(x) and q(x) be polynomials and c a real number. Then:
1
2
lim p(x) = p(c)
x→c
lim
x→c
p(x) p(c)
=
, where q(c) , 0
q(x) q(c)
Remark
So polynomials are continuous everywhere and rational functions are
continuous at every point in their domain. The next theorem lists
other common functions that are either continuous everywhere or
continuous at all points in their respective domains.
1.3 Finding limits analytically
Other Limit Laws
Theorem
Let c be a real number in the domain of the given function, let n be a
positive integer, and let a > 0. Then the following limits hold:
1
2
3
lim sin x = sin c
4
lim cos x = cos c
5
lim tan x = tan c
6
x→c
x→c
x→c
lim csc x = csc c
7
lim sec x = sec c
8
lim cot x = cot c
9
x→c
x→c
x→c
Example
Evaluate the following limits:
1
lim cos x =
3
2
lim csc2 x − cot2 x =
x→3
lim ln e3x =
x→0
x→π/3
4
lim e
x→1
√
4x
=
lim ax = ac
x→c
lim ln x = ln c
√
√
lim n x = n c
x→c
x→c
1.3 Finding limits analytically
Other Limit Laws
Theorem
Let c be a real number in the domain of the given function, let n be a
positive integer, and let a > 0. Then the following limits hold:
1
2
3
lim sin x = sin c
4
lim cos x = cos c
5
lim tan x = tan c
6
x→c
x→c
x→c
lim csc x = csc c
7
lim sec x = sec c
8
lim cot x = cot c
9
x→c
x→c
x→c
lim ax = ac
x→c
lim ln x = ln c
√
√
lim n x = n c
x→c
x→c
Example
Evaluate the following limits:
1
lim cos x = 1/2
3
2
lim csc2 x − cot2 x = 1
x→3
lim ln e3x = 0
x→0
x→π/3
4
lim e
x→1
√
4x
= e2
1.3 Finding limits analytically
Squeeze theorem
Theorem
(Squeeze theorem) Let f , g and h be functions on an open interval I
containing c such that for all x in I we have f (x) ≤ g(x) ≤ h(x). If
lim f (x) = L = lim h(x)
x→c
x→c
then
lim g(x) = L
x→c
Example
sin x
=1
x→0 x
(A fundamental trig limit) lim
Prove!
Remark
This says that sin x and x are approaching 0 at the same rate.
1.3 Finding limits analytically
Special limits
Theorem
1
2
sin x
=1
x→0 x
cos x − 1
= 0 Prove!
lim
x→0
x
lim
3
lim (1 + x)1/x = e
x→0
4
ex − 1
=1
x→0
x
lim
Remark
So, cos x − 1 approaches 0 faster than x does, while ex − 1 and x
approach 0 at the same rate. These are all examples of indeterminate
forms, which we will study in more detail in a later section (§6.7).
1.3 Finding limits analytically
Special limits
Example
Find the following limits:
1
2
sin(3x)
=
x→0
x
sin(5x)
lim
=
x→0
8x
lim
3
4
tan(4x)
=
x→0
3x
lim
lim x csc(2x) =
x→0
1.3 Finding limits analytically
Special limits
Example
Find the following limits:
1
2
sin(3x)
=3
x→0
x
sin(5x)
lim
= 5/8
x→0
8x
lim
3
4
tan(4x)
= 4/3
x→0
3x
lim
lim x csc(2x) = 1/2
x→0
1.3 Finding limits analytically
Special limits
Remark
In a rational function, if numerator and denominator each contain a
factor that is making the denominator zero, canceling the common
factor can make it possible to evaluate the limit.
Example
x2 + x − 6
.
x→2
x−2
Find lim
1.3 Finding limits analytically
Using algebra to find limits
Theorem
Let g(x) = f (x) for all x in an open interval, except possibly at c, and
let lim g(x) = L. Then
x→c
lim f (x) = L
x→c
Example
Find the following limits:
1
2
3
x2 + 3x
=
x→0 x2 − 2x
x2 + 9x + 8
lim 2
=
x→−1 x − 6x − 7
x2 − 4x − 12
lim 2
=
x→6 x − 13x + 42
lim
4
5
6
1
x
− 12
lim
=
x→2 x − 2
x3 − 7x + 6
lim 3
=
x→2 x − 4x2 + x + 6
x4 − 2x3 − 7x + 14
lim
=
x→2
x4 − 16
1.3 Finding limits analytically
Using algebra to find limits
Theorem
Let g(x) = f (x) for all x in an open interval, except possibly at c, and
let lim g(x) = L. Then
x→c
lim f (x) = L
x→c
Example
Find the following limits:
1
2
3
x2 + 3x
= −3/2
x→0 x2 − 2x
x2 + 9x + 8
lim 2
= −7/8
x→−1 x − 6x − 7
x2 − 4x − 12
lim 2
= −8
x→6 x − 13x + 42
lim
4
5
6
1
x
− 12
lim
= −1/4
x→2 x − 2
x3 − 7x + 6
lim 3
= −5/3
x→2 x − 4x2 + x + 6
1
x4 − 2x3 − 7x + 14
lim
=
4
x→2
32
x − 16
1.3 Finding limits analytically
Just checking. . . .
1 You are given the following information:
• lim f (x) = 0 = lim g(x)
x→1
x→1
• lim f (x)/g(x) = 2.
x→1
2
3
4
What can be said about the relative sizes of f (x) and g(x) as x
approaches 1?
5x
Find lim
.
x→0 cos(3x)
5x
.
Find lim
x→0 sin(3x)
Find lim cos x sin x.
x→π/4
x2 − 5x − 14
.
x→−2 x2 + 10x + 16
5
Find lim
6
If f (x) = x2 − x, find lim
h→0
f (4 + h) − f (4)
.
h
1.4 One-sided limits
Left- and right-hand limits
Definition
Let f be a function defined on an open interval containing c. The
notation
lim− f (x) = L,
x→c
which is read as “the limit of f (x) as x approaches c from the left is L,”
or “the left-hand limit of f at c is L,” means that:
given any ε > 0, there exists δ > 0 such that whenever
0 < c − x < δ, we have |f (x) − L| < ε.
Remark
What changes need to be made to define a right-hand limit at c?
1.4 One-sided limits
Left- and right-hand limits
Example
The graph of the function f (x) is shown below at right. Based on the
graph, determine the following:
1
2
3
lim f (x) =
10
lim f (x) =
11
lim f (x) =
12
x→−1−
x→−1+
x→−1
f (−1) =
5 lim f (x) =
−
4
x→2
6
7
8
9
lim f (x) =
x→2+
lim f (x) =
x→2
f (2) =
lim f (x) =
x→0
f (0) =
lim f (x) =
x→4
f (4) =
1.4 One-sided limits
Left- and right-hand limits
Example
The graph of the function f (x) is shown below at right. Based on the
graph, determine the following:
1
2
3
lim f (x) = 2
10
lim f (x) = 2
11
lim f (x) = 2
12
x→−1−
x→−1+
x→−1
f (−1) = 2
5 lim f (x) = 8
−
4
x→2
6
7
8
9
lim f (x) = 4
x→2+
lim f (x) = dne
x→2
f (2) = 2
lim f (x) = 0
x→0
f (0) = 4
lim f (x) = 2
x→4
f (4) = dne
1.4 One-sided limits
Left- and right-hand limits
Remark
From this example, we learn two important things about limits. First,
lim f (x) and f (c) are independent of one another:
x→c
lim f (x) can be f (c), but it need not be; in fact, lim f (x) can exist
x→c
even if f (c) doesn’t exist, and vice versa.
x→c
The second important thing we learn from this example is . . .
Theorem
Let f be a function defined on an open interval I containing c. Then
lim f (x) = L ⇐⇒ lim− f (x) = L and
x→c
x→c
lim f (x) = L
x→c+
1.4 One-sided limits
Just checking. . . .
1
True or false: if lim− f (x) = 3, then lim f (x) = 3.
2
True or false: if lim f (x) = 3, then lim− f (x) = 3.
3
True or false: if lim f (x) = 3, then f (5) = 3.
x→5
x→5
x→5
x→5
x→5
4
5
x2 + 5.8x − 1.2
x→0.2 x2 − 4.2x + 0.8
2
x − 6x + 9
Evaluate the limit: lim
x→3 x3 − 3x
Estimate the limit numerically: lim
1.5 Continuity
Definition
Let f be a function defined on an open interval I containing c.
1
f is continuous at c if lim f (x) = f (c).
2
f is continuous on I if f is continuous at c for all values of c in I.
3
If f is continuous on (−∞, ∞), we say f is continuous
everywhere.
x→c
Definition
Let f be defined on the closed interval [a, b] for some real numbers
a, b. f is continuous on [a, b] if:
1
2
3
f is continuous on (a, b),
lim f (x) = f (a), and
x→a+
lim f (x) = f (b).
x→b−
1.5 Continuity
Example
Determine the domain of the given function and the interval(s) on
which it is continuous.
Function
f (x) = 1/(x − 2)
f (x) =
√
x−1
f (x) =
√
4 − x2
(
f (x) =
−3x + 4, x < 2
2x − 1, x ≥ 2
Domain
Interval(s) of continuity
1.5 Continuity
Example
Determine the domain of the given function and the interval(s) on
which it is continuous.
Function
f (x) = 1/(x − 2)
Domain
Interval(s) of continuity
(−∞, 2) ∪ (2, ∞)
(−∞, 2), (2, ∞)
f (x) =
√
x−1
[1, ∞)
[1, ∞)
f (x) =
√
4 − x2
[−2, 2]
[−2, 2]
(−∞, ∞)
(−∞, 2), [2, ∞)
(
f (x) =
−3x + 4, x < 2
2x − 1, x ≥ 2
1.5 Continuity
Properties of continuous functions
Theorem
Let f and g be continuous functions on an interval I, let c be a real
number, and let n be a positive integer. The following functions are
continuous on I.
1
f ±g
2
c·f
3
f ·g
4
f /g (g , 0 on I)
5
fn
p
n
6
7
f
If the range of f is J and g is continuous on J, then g ◦ f is
continuous on I.
1.5 Continuity
Common continuous functions
Theorem
The following functions are continuous on their domains.
1
f (x) = sin x
6
f (x) = cot x
2
f (x) = cos x
7
f (x) = ln x
3
f (x) = tan x
8
4
f (x) = csc x
9
5
f (x) = sec x
f (x) = ax (a > 0)
√
f (x) = n x
(n a positive integer)
Example
Find the interval(s) on which the following functions are continuous.
√
√
1 f (x) =
3 − x + 2x − 1 is continuous on
√
2 f (x) =
ln(2x + 3) is continuous on
1.5 Continuity
Common continuous functions
Theorem
The following functions are continuous on their domains.
1
f (x) = sin x
6
f (x) = cot x
2
f (x) = cos x
7
f (x) = ln x
3
f (x) = tan x
8
4
f (x) = csc x
9
5
f (x) = sec x
f (x) = ax (a > 0)
√
f (x) = n x
(n a positive integer)
Example
Find the interval(s) on which the following functions are continuous.
√
√
1 f (x) =
3 − x + 2x − 1 is continuous on [1/2, 3]
√
2 f (x) =
ln(2x + 3) is continuous on [−1, ∞)
1.5 Continuity
Intermediate value theorem
Theorem
(Intermediate value theorem) Let f be a continuous function on [a, b].
Then for every value y-value in between f (a) and f (b), there is an
x-value c in [a, b] such that f (c) = y.
Remark
The IVT is an existence theorem: it asserts the existence of an x-value
c with f (c) = y for every y in between f (a) and f (b). We can find a
good approximation of c using the bisection method, as illustrated in
the next example.
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
Interval
Midpoint sign
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
Interval
[−1, 0]
Midpoint sign
f (−0.5) > 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
Interval
[−1, 0]
[−1, −0.5]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
4
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
[−0.75, −0.625]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
4
5
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
[−0.75, −0.625]
[−0.6875, −0.625]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
f (−0.65625) > 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
4
5
6
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
[−0.75, −0.625]
[−0.6875, −0.625]
[−0.6875, −0.65625]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
f (−0.65625) > 0
f (−0.67188) > 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
4
5
6
7
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
[−0.75, −0.625]
[−0.6875, −0.625]
[−0.6875, −0.65625]
[−0.6875, −0.67188]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
f (−0.65625) > 0
f (−0.67188) > 0
f (−0.67969) > 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
1
2
3
4
5
6
7
8
Interval
[−1, 0]
[−1, −0.5]
[−0.75, −0.5]
[−0.75, −0.625]
[−0.6875, −0.625]
[−0.6875, −0.65625]
[−0.6875, −0.67188]
[−0.6875, −0.67969]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
f (−0.65625) > 0
f (−0.67188) > 0
f (−0.67969) > 0
f (−0.68359) < 0
1.5 Continuity
Example
(Bisection method) Find the root of f (x) = x3 + x + 1 accurate to two
decimal places. Note, f (−1) = −1 < 0 and f (0) = 1 > 0.
Iteration
Interval
1
[−1, 0]
2
[−1, −0.5]
3
[−0.75, −0.5]
4
[−0.75, −0.625]
5
[−0.6875, −0.625]
6
[−0.6875, −0.65625]
7
[−0.6875, −0.67188]
8
[−0.6875, −0.67969]
9
[−0.68359, −0.67969]
Midpoint sign
f (−0.5) > 0
f (−0.75) < 0
f (−0.625) > 0
f (−0.6875) < 0
f (−0.65625) > 0
f (−0.67188) > 0
f (−0.67969) > 0
f (−0.68359) < 0
1.5 Continuity
Just checking. . . .
1
True or false: If f is defined on an open interval containing c and
lim f (x) exists, then f is continuous at c.
x→c
2
True or false: If f is continuous at c, then lim f (x) exists.
3
True or false: If f is continuous at c, then lim f (x) = f (c).
4
5
x→c
x→c
True or false: If f is continuous on [0, 1) and on [1, 2], then f is
continuous on [0, 2].
( 2
)
x
x≤2
Let f (x) =
. Find the value of m that makes f
3 − mx x > 2
continuous at x = 2.
1.6 Limits involving infinity
Infinite limits
Definition
We say lim f (x) = ∞ if for every M > 0 there exists δ > 0 such that if
x→c
0 < |x − c| < δ, then f (x) ≥ M
Example
1
2
1
=∞
x→3 (x − 3)2
lim − tan x = ∞ and lim + tan x = −∞.
lim
x→π/2
x→π/2
Definition
If the limit of f (x) as x approaches c from either the right or the left
(or both) is ∞ or −∞, we say that the function has a vertical
asymptote at c.
1.6 Limits involving infinity
Infinite limits
Example
1
2
x2
has vertical asymptotes at x = 1 and x = −1.
x2 − 1
x3 + 2x2 − x − 2
f (x) =
does not. Indeed, f (x) = x + 2 for all x
x2 − 1
except x = 1 and x = −1, where is simply has a hole (i.e. point
discontinuity).
f (x) =
Remark
The above example illustrates the fact that just because a function has
a denominator of zero for particular values of x does not mean that it
has a vertical asymptote at those values of x. It must also have an
infinite limit (from the left, or from the right, or both) in order to have
a vertical asymptote.
1.6 Limits involving infinity
Limits at infinity
Definition
1
2
3
We say that lim f (x) = L if for every ε > 0 there exists M > 0
x→∞
such that if x ≥ M, then |f (x) − L| < ε.
We say that lim f (x) = L if for every ε > 0 there exists M < 0
x→−∞
such that if x ≤ M, then |f (x) − L| < ε.
If lim f (x) = L or lim f (x) = L, we say that y = L is a
x→∞
x→−∞
horizontal asymptote of f .
Example
Find the horizontal asymptote(s) of the following functions.
1
f (x) =
x2
x2 − 1
2
f (x) = √
3x
x2 + 1
3
f (x) =
sin x
x
1.6 Limits involving infinity
Limits at infinity
Definition
1
2
3
We say that lim f (x) = L if for every ε > 0 there exists M > 0
x→∞
such that if x ≥ M, then |f (x) − L| < ε.
We say that lim f (x) = L if for every ε > 0 there exists M < 0
x→−∞
such that if x ≤ M, then |f (x) − L| < ε.
If lim f (x) = L or lim f (x) = L, we say that y = L is a
x→∞
x→−∞
horizontal asymptote of f .
Example
Find the horizontal asymptote(s) of the following functions.
1
x2
f (x) = 2
x −1
y=1
2
f (x) = √
y = ±3
3x
x2
+1
3
f (x) =
y=0
sin x
x
1.6 Limits involving infinity
Limits at infinity
Theorem
Suppose we have a rational function
f (x) =
an xn + · · · + a1 x + a0
bm xm + · · · + b1 x + b0
where any of the coefficients may be zero except for an and bm .
an
1 If n = m, then lim f (x) =
x→±∞
bm
2 If n < m, then lim f (x) = 0.
x→±∞
3
If n > m, then lim f (x) and lim f (x) are both infinite.
x→∞
x→−∞
1.6 Limits involving infinity
Limits at infinity
Example
Find the following limits.
1
2
6x − 2x2
=
x→∞ x2 − 2x − 3
6x − 2x3
lim 2
=
x→−∞ x − 2x − 3
lim
3
4
6x − 2x2
=
x→∞ x3 − 2x − 3
6x − 2x2
lim 2
=
x→3 x − 2x − 3
lim
1.6 Limits involving infinity
Limits at infinity
Example
Find the following limits.
1
2
6x − 2x2
= −2
x→∞ x2 − 2x − 3
6x − 2x3
lim 2
=∞
x→−∞ x − 2x − 3
lim
3
4
6x − 2x2
=0
x→∞ x3 − 2x − 3
6x − 2x2
lim 2
= −3/2
x→3 x − 2x − 3
lim
1.6 Limits involving infinity
Just checking. . . .
1
True or false: If lim f (x) = ∞, then f has a vertical asymptote at
x = 3.
x→3
2
True or false: If lim f (x) = ∞, then f (3) is not defined at x = 3.
3
True or false: If lim f (x) = ∞, then f is not continuous at x = 3.
4
Using a ε − δ argument, show that lim 3x − 1 = 2.
x→3
x→3
x→1
5
Evaluate the limit lim
x→−4 x2
x2
− 16
.
− 4x − 32