MATH 151 Engineering Math I, Spring 2014
JD Kim
Week3 Section 2.2, 2.3, 2.5-2.6, (quantitative) definition of limit, calculation of
limits, limits at infinity, continuity.
Section 2.2 The limit of a function
Let’s investigate the behavior of the funtion f defined by f (x) = x2 − x + 2 for
values of x near 2. The following table gives values of f (x) for values of x close to 2
but not equal to 2.
x
f (x)
x
f (x)
1.0
2.000000
3.0
8.000000
1.5
2.750000
2.5
5.750000
1.8
3.440000
2.2
4.640000
1.9
3.710000
2.1
4.310000
1.95
3.852500
2.05
4.152500
1.99
3.970100
2.01
4.030100
1.995
3.985025
2.005
4.015025
1.999
3.997001
2.001
4.003010
From the table above, we see that when x is close to 2 (on either side of 2), f (x)
is close to 4. In fact, it appears that we can make the value of f (x) as close as we
like to 4 by taking x sufficiently close to 2.
We express this by saying ”the limit of the function f (x) = x2 −x+2 as x approaches
2 is equal to 4”.
The notation for this is
lim (x2 − x + 2) = 4
x→2
1
Definition. 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 but not equal to a.
This says that the value of f (x) get closer and closer to the number L as x gets
closer and closer to the number a (from either side of a), but x 6= a.
Ex1) limx→1
x−1
=?
x2 − 1
undefined at x = 1, but there is limit, because the definition of limx→a f (x) says
that we consider values of x that are close to a but not equal a.
x<1
f (x)
x>1
f (x)
0.5
0.666667
1.5
0.400000
0.9
0.526316
1.1
0.476190
0.99
0.502513
1.01
0.497512
0.999
0.500250
1.001
0.499750
0.9999 0.500025
1.0001 0.499975
From the table and figure, we make the guess that the equation approaches 0.5 as x
approaches 1.
2
Ex2) Let f (x) =



 2


 x
x=1
, find limx→1 f (x).
otherwise
Note, this function is not continuous.
3
Definition. A function f is continuous at a number a if
lim f (x) = f (a)
x→a
or lim f (x) = lim f (x) = f (a)
x→a+
Ex3) Let f (x) =




2−x







 x



4







 4−x
x→a−
if x < −1
if − 1 ≤ x < 1
,
if x = 1
if x > 1
sketch the graph of f (x) and find all values of x for which the limit does not exist.
In sum,
lim f (x) = L if and only if
x→a
lim f (x) = L and
x→a−
Infinite Limits and Vertical asymtotes.
Let’s start from f (x) =
1
.
x
4
lim f (x) = L
x→a+
Ex4) f (x) =
1
, find limx→0 f (x).
x2
Ex5) Find the limits;
5-1) limx→5−
6
x−5
5-2) limx→5+
6
x−5
5-3) limx→5
6
x−5
5-4) limx→0−
5-5) limx→0+
5-6) limx→0
x−1
+ 2)
x2 (x
x−1
+ 2)
x2 (x
x−1
+ 2)
x2 (x
5
x+1
. For each vertical
− 2x − 3
asymtote, describe the behavior of f (x) near the asymtote.
Ex6) Find the vertical asymtotes for f (x) =
6
x2
Section 2.3.
laws
Calculating limits using the limit
Limit Laws Suppose that c is a constant and the limits
lim f (x) and lim g(x)
x→a
x→a
exist. Then
1.
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
2.
lim [f (x) − g(x)] = lim f (x) − lim g(x)
x→a
x→a
x→a
3.
lim [cf (x)] = c lim f (x)
x→a
x→a
4.
lim [f (x) · g(x)] = lim f (x) · lim g(x)
x→a
5.
x→a
x→a
limx→a f (x)
f (x)
=
if lim g(x) 6= 0
x→a
x→a g(x)
limx→a g(x)
lim
6.
lim [f (x)]n = [lim f (x)]n
x→a
x→a
7.
lim c = c
x→a
8.
lim x = a
x→a
9.
lim xn = an
x→a
10.
lim
x→a
√
n
7
x=
√
n
a
11.
lim
x→a
p
n
f (x) =
p
n
f (a)
where n is a positive integer. (If n is even, we assume that limx→a f (x) > 0).
Ex7) Find limx→−2
x3 + 2x2 − 1
.
5 − 3x
Ex8) Find limx→−2(x2 + x + 1)5
Ex9) Find limx→4−
√
Ex10) Find limx→4+
16 − x2
√
16 − x2
8
Ex11) Find limx→1
x4 + x2 − 6
x4 + 2x + 3
Ex12) Find limx→−3
Ex13) Find limx→9
x2 − x − 12
x+3
x2 − 81
√
x−3
Ex14) Find limx→3 √
x−3
x2 + 7 − 4
9
Ex15) Find limx→2
x−4
(x − 2)2
Ex16) If it ispknown that limx→2 f (x) = 3, find
f (x) + 6
16-1) limx→2
f (x) + 2x + 11
16-2) limx→2 (f (x))2
Ex17) limt→1
1−t
, 2t
t2 − 1
10
Ex18) Show that limx→0 |x| = 0
Ex19) Prove that limx→0
|x|
does not exist.
x
11
Ex20) Let f (x) =
x2 + 3x
, find limx→−3 f (x).
|x + 3|
The Squeeze Theorem If f (x) ≤ g(x) ≤ h(x) for all x in open interval that
contains a (except possibly at a) and
lim f (x) = lim h(x) = L
x→a
x→a
then
lim g(x) = L
x→a
12
Ex21) Show that limx→0 x sin
1
= 0.
x
13
Section 2.5 Continuity
Definition A function f is continuous at a number a if
lim f (x) = f (a)
x→a
Note Definition implicitly requires three things if f is continuous at a;
1. f (a) is defined (that is, a is in the domain of f and has a function value at a)
2. limx→a f (x) exists, i.e., limx→a+ f (x) = limx→a− f (x).
3. limx→a f (x) = f (a)
Polynomial is always continuous
Ex22) Show that f (x) = x2 − 3x + 2 is continuous at any point.
14
In each case the graph cannot be drawn without lifting the pen from the paper,
because a hole or break or jump occurs in the graph.
Definition The kind of discontinuity illustrated first three is called removable
becuase we could remove the discontinuity by redefining f at 2.
The discontinuity in second figure is called an infinite discontinuity.
The discontinuityies in the last figure are called jump discontinuities because the
function ”jumps” from one value to another.
Definition We say f (x) is continuous from the right at x = a if limx→a+ f (x) =
f (a). similarly, we say f (x) is continuous from the left at x = a if limx→a− f (x) =
f (a).
15
Ex23) From the accompaning figure, state the numbers at which f is discontinuous. For each of the numbers stated, state whether f is continuous from the right,
or from the left, or neither.
Ex24) Explain why the following functions are not continuous at the indicated
values of x.
24-1) f (x) =
−1
, x=1
(1 − x)2
16
24-2) f (x) =



 2x + 1 if x ≤ 0


 3x
, x=0
if x > 0


x2 − 2x − 8


if x 6= 4
x−4
, x=4
24-3) f (x) =


 3
if x = 4
17
Ex25) Find all points of discontinuity for




2x + 1 if







 3x
if
f (x) =



x2 + 2 if







 4
if
18
x < −1
−1<x<1
x>1
x=1
Ex26) If g(x) =
continuous?



 x2 − c2
if x < 4


 cx + 20 if x ≥ 4
19
. For what value(s) of c make(s) g(x)




x2 + c if x > 1




Ex27) If f (x) =
4
if x = 1 . For what value(s) of c is f (x) continuous,






 4cx + 3 if x < 1
if any?
20
Ex28) Which of the following functions has removable discontinuity at x = a? If
the discontinuity is removable, find a function g that agrees with f for x 6= a and is
continuous at x = a.
28-1) f (x) =
x4 − 16
, x = 2.
x−2
28-2) f (x) =
1
, x = 1.
x−1
28-3) f (x) =



 x2
if x < 1
, x=1


 2x + 4 if x ≥ 1
21
Intermediate Value Theorem
Suppose that f is continuous on the closed interval [a, b] and let N be any number
strictly between f (a) and f (b). Then there exists a number c in (a, b) such that
f (c) = N.
2.
Ex29) Show there is a root of the equation 4x3 − 6x2 + 3x − 2 = 0 between 1 and
Ex30) If g(x) = x5 − 2x3 + x2 + 2, show there a number c so that g(c) = −1.
22