# 12.1 limits by substitution

```egy that would
eventually
work),
but these
concepts
were763
not
well Heine
understood
until
6965_CH10_pp735-778.qxd
1/20/10
3:35
PMhis Page
Karl
(1815–1897)
and
student
Eduard
(1821–1881)
introduced
Karl Weierstrass
(1815–1897)
andWeierstrass
his student
Eduard Heine
(1821–1881)
introduced
Karl Weierstrass (1815–1897) and his student Eduard Heine (1821–1881) introduced
the formal,
unassailable
definitions
that are used
in ourtohigher mathematics courses tothe formal, unassailable definitions
that
are
usedPage
in our
higher mathematics
courses
6965_CH10_pp735-778.qxd
3:35
PM
the formal, 1/20/10
unassailable
definitions
that763
are used in our higher mathematics courses today.
By that
time,
Newton
and
Leibniz
day. By that time,
Newton
and
Leibniz
been
for
over
150
years.
day. By that time, Newton and Leibniz had been dead for over 150 years.
Excelsior PreCalc Honors
Name____________________________
12.1 – Evaluating Limits Numerically by Substitution
6965_CH10_pp735-778.qxd
PM 763
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6965_CH10_pp735-778.qxd
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Page
a Limit Informally
DefiningDefining
a LimitDefining
Informally
a Limit
the limitInformally
of each of the following functions:
S
Warm up
There
nothing
difficult
the following
limit statements:
There is nothing
difficult
the is
following
statements:
There
is nothing
difficult
the following
statements:
SECTION 10.3 More o
1
1
2 12x - 12q= 25
lim
1x 20 + lim
32 =1 q
lim
= 0
lim
= 5 12x -lim
1x
+ 32 lim
= 1x + lim
=
lim 12x - 12 lim
q
12
=
5
32
=
=
0
q
q x:3
q n
n: q n
x:3
x:SECTION
n:x:
EXERCISES
x:3
x : q 10.3
n: q n
SECTION
10.3
EXERCISES
That
is
why
have
usedbackground
limit
notation
throughout
this
Particularly
when
That is why we
have
used
limit
notation
throughout
this
book.
Particularly
when
That
is why
we
have
usedwe
limit
throughout
this
book.
Particularly
Exercise
numbers
with
anotation
gray
indicate
problems
thatbook.when
29. (a) lim
ƒ1x2
x
:
3graphers
the limiting
behavioralgeof functions algeelectronic graphers
are available,
analyzing
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algetheelectronic
authors
designed
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abehavior
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electronic
graphers
arehave
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the analyzing
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of functions
numbers
with
a graysubstitution:
background indicate problems that
29.much
(a) lim
ƒ1x2 we need
ƒ1x2 y
(b)to lim
Example
1 Exercise
Find
the
limitand
by
direct
braically,
numerically,
and
graphically
canneed
tellifwhat
us
of -what
know
braically,
numerically,
graphically
can
telltheuslimit
much
of
we
to
know
braically,
numerically,
and
graphically
can
tellwhat
ussubstitution
much
of
we need
to know
x:3
x:
3+
In Exercises
1–10,
find
by
direct
it exists.
the authors have designed
to be solved
without
a calculator.
(b) lim+ ƒ1x2
(c) lim ƒ1x2
12
lim
1x -substitution
12
2. lim 1x - 12
In Exercises
1–10, find the1.limit
by xdirect
if b.
it exists.
a.
x : -1
x:3
x:3
x:3
4
3
What
difficult
is to
with
an air-tight
definition
what
a(c)
limit
If really is. If 2
ƒ1x2
What
is come
difficult
is
to
come
upofwith
anaair-tight
definition
ofreally
whatis.
a limit
What is difficult
is toiscome
up with
an
air-tight
definition
what
limitofreally
is.
If lim
3 up
x:3
- 1212
1. lim x 1x - 122 3. lim 1x2.
- lim
2x 1x
+ 32
4. lim 1x 3 - x + 52
been easy,
itx :
would
not
taken
150
years.
subtleties
of the
“epsilonx:
-1
x:3
been
easy,
ityears.
would
not
have
years.
The
subtleties
of the “epsilonit had been easy,
it would
not
have
150have
The
subtleties
of150
the
“epsilon2taken
x :taken
-2 The
1
3
3
2/3
(a) limbut
ƒ1x20
lim
1x
2x
+
32
lim
1x
x
+
52
3.
4.
delta”
definition
of
Weierstrass
and
Heine
are
as
beautiful
as
they
are
profound,
but
lim
lim
1x + are
5 ofasWeierstrass
1x -profound,
42 as beautiful
5. and
delta”
definition
and
Heine
are
profound,
delta” definition
of Weierstrass
Heine
beautiful 6.
as
they
are
but as they are30.
6965_CH10_pp735-778.qxd
1/20/10
Page
x : 1x:2
x: -23:35 PM
x:2
x : -2 763
they
are not the
stuff
of acourse.
precalculus
course.
Therefore,
even
as
welimlook
more
they are not
stuff
precalculus
even
as course.
wePMlook
more
are not
of42a2/3
precalculus
Therefore,
even
more
6965_CH10_pp735-778.qxd
3:35
Page
(a) 763
ƒ1x2 as we look
(b) lim
ƒ1x2 y
limstuff
lim
1x +of5 a they
1x Therefore,
-1/20/10
5. the
6. the
x 30.
+
x properties
x:1
x
:
1
x:2
x:
-2
closely
at
limits
and
their
in
this
section,
we
will
continue
to
refer
to
our
lim
lim
1e
sin
x2
ln
a
sin
b
7.
8.
Exercise
numbers
with
a
gray
background
indicate
problems
that
closely at limits and their properties
this section,
will continue
to section,
refer to we
our will continue to refer to our 29.
closely
atinlimits
and theirweproperties
in this
x:0
x:p
2 We (b)
limit ƒ1x2
ƒ1x2 4
(c) lim for
x that
“informal”
definition
of
limit
(essentially
of
d’Alembert).
here for
xthe
“informal” 7.
definition
of
limit
(essentially
that
of
d’Alembert).
We
repeat
it
here
for
“informal”
definition
of
limit
(essentially
that
of
d’Alembert).
x:1 + We repeat it here
authors
have
designed
without
a repeat
calculator.
x:1
lim lnbackground
lim 1e
sin x2
asin to
b be solved
8.gray
Exercise
numbers
with
a
indicate
problems
that
2
29.
(a)2
x:0
x:p
2
x - 1
reference:
(c) lim ƒ1x2
9. lim 1x 2 - 22
10. lim
1
SECTION 10.3 EXERCISES
SECTION 10.3 EXERCISES
x:1
the authorsInhave
designed
be
solved
asubstitution
Exercises
find
the
limit without
byx :
direct
if
it exists.
x : a1–10, to
a x2 +
1calculator.
x2 - 1
0
(b)the
In Exercises 31 and 32,
9. lim 1x 2 - 22
10. lim 2 2
12
In1.
Exercises
11–18,
(a)
explain
why
you
cannot
use
substitution
to
find
x:a
x:a
x by
+ 1direct substitution
limsubstitution:
x 1xlimit
- 12
2. lim 1xif-it 12
Which of the statements ab
InFind
Exercises
find
the
exists.
You Try
the limit1–10,
by
thedirect
limit
and
limit at
algebraically
if itx:3
exists.In Exercises 31 and 32, the graph of a function y =
x:
-1 (b) find theLimit
(INFORMAL)
a
DEFINITION
(INFORMAL)
Limit
at
a
In ExercisesDEFINITION
11–18, (a) explain
why
you
cannot
use
substitution
to
find
DEFINITION
(INFORMAL) Limit at a2 Which
(c)
lim are
ƒ1x2
31.the
(a)function
true=
x : -1 +
x32 + =
7x
12
x 12
- 91x 3 - close
lim
1x2ƒ1x2
2x
32mean
lim
x + to52L
3.
4. lim
limfind
xthe
1x
12
1x
1.andWhen
2.
the limit
(b)
limit
algebraically
ifL+it,”+exists.
lim
we
write
“
we
that
ƒ1x2
gets
arbitrarily
lim
lim
11.
12.
a.
b.
L,”
When we write x:
“ lim
we
that
ƒ1x2
gets=x:3
arbitrarily
close
to 31.
Lƒ1x2(a)
x:2
-2
ƒ1x2
L,” we mean
write
“ lim
that
getslim
arbitrarily
to L lim ƒ1x2 =
-1ƒ1x2 =When
x:
awemean
ƒ1x2 =close
1 (b)
2x:
2
x 2 -not
9x:a
x:
-1 +
+ 7x
+ 12 x : -3
x 2 - 9to a. x : 3 x 3+ 2x - 15
x : 0as xx:a
gets
arbitrarily
close (but
equal)
2/3
3
30.
as x gets11.
arbitrarily
(but
not
equal)
to
a.
lim
limlimclose
12.
lim
lim
1x
+
5
1x
42
5.
6.
as
x
gets
arbitrarily
close
(but
not
equal)
to
a.
3
3
2
1x
2x
+
32
lim
1x
x
+
52
3.
4.
2
x +x:3
1 x 2 + 2x - 15
x x:
- 2x
+ x -(b)2 lim- ƒ1x2 = 0
(c) lim- ƒ1x2 = 1
x: -3
x
9
x:2
-2
x:0
x:2
x: -2
x:0
lim
13. lim
14.
x : -1 x + 1
x - 2/3
2 (c) lim ƒ1x2 = 1
x3 + 1
x 3 - 2x 2 + x - x2: 2
lim
ƒ1x2
(d)
x
30.
(a)=
x2 lim
+lim
5 1e14.
6. lim 8.1xƒ xlim
x:0 2- 42
13.5.limlim 1x
x:
0sin
x2
ln
asin
b
7.
4
ƒ
x
4
x: -1
x:2
x
+
1
x
2
x:2
x: -2
(d)2 lim- ƒ1x2 = lim+ ƒ1x2
x:0
lim
lim x:p
15.
16.
(e) lim ƒ1x2 exist
2
x:0
x:0
x : -2 x + 2
x : -2 x + 2
x:0
2
x
4
ƒ
ƒ
x - 4
(b)ƒ1
x
x Exercise
limbackground
15. lim numbers
16.
lim=
lim problems
ƒ1x2 exists 29.
(e)that
lim (f)
ƒ1x2
1
(g)
2
Exercise
with
a
gray
indicate
problems
numbers
with
a
gray
background
indicate
that
29
lim
(a)
ƒ1
lim
7.
- 2 x b- 1 x:0
x: -2
x: -2 x + 28. lim lnx asin
x:0
x + 1e
2 sin x2
x:0
2 - 3
lim
lim
1x
17.
18.
x:3
x:0
x:p
2
lim
-have
22 solved
lim
9.the
2
designed
to x10.
be
solved
without
the authors have
designed
be
without
calculator.
xauthors
: 01x to
: 0a
lim ƒ1x2 = 1
(g)a calculator.
ƒ1x2lim
= ƒ1
0
(i) lim (h)
(c)
xx:a
x - 2
x:a
x:1 ƒ1
x(b)
:1
x 2 + 1 x:0
lim
17. lim 1x - 3
18. lim
+
sin
2x direct
x:0
x:0
x:3
lim
ƒ1x2
= 0 32. (a) lim(j)
ƒ1
(i) lim if
Infind
Exercises
1–10,
limit
by
it exists.
In
Exer
ƒ1x2
x 2find
In Exercises
1–10,
the
limit
by
direct
itsubstitution
exists.
-=
1, along
lim x
=cannot
1if
In2 Exercises
19–22,
use
the
fact the
thatsubstitution
withsubstitution
the
x:1
x : -1 + x:2
In
Exercises
11–18,
(a)
explain
why
you
use
to
find
x
:
0
x
9. lim 1x - 221.2 lim sin
lim ƒ1x
2 lim
12 ƒ1x2 = 1 (b) (c)
x - 10.
Which
12 lim 32.
lim
lim ƒ1x2
x the
1x
1x (a)
- 12
2 122.
+
lim
= following
12.
In Exercises
19–22,
use
fact12
that
, 12
along
with
the
x:3does
lim
lim
xlimit
1xtheproperties,
-and
1x
1. x:a
x:a
limit
to
find
the
limits.
x:
-1
the
(b)
find
limit
algebraically
if
it
exists.
x
:
2
x
+
1
x:
-1 x
x:3
x:0
x: -1
x:3
ƒ1x2 31.
= 2
(c)
Inxlim
Exercises
3
3 lim ƒ1x2 does not exist
sin
x
sin
3x
limit properties, to find3 the following
limits.
2
2 (b)
3
lim
1x7xwhy
-+2x
+ lim
32
lim
1x
-x:2
x to
+ find
52
3.+explain
4.
:2
xx:2
+
12
x
9
1x 19.
- 2x
32
1x
x
+
52
3. lim 11–18,
4.
lim
lim
20.
In Exercises
(a)
you
cannot
use
substitution
x : 0 2x 2 - x
:0
xlim x: -2 (c) lim ƒ1x2 = 2
Which
of the
x:2
lim ƒ1x2
=
(d)
sin x 11. lim
sin 3x x: -2 x12.
2/3
2
2 1x - x:2
x : 1x:5.
-3 20.
x:3
30
the19.
limit
(b)
find
the
limit
algebraically
if
it
exists.
lim
lim and
2/3
lim
lim
1x
+
5
42
6.
2
x
9
x
+
2x
15
2 1x + 5
(a) ƒ1x2
lim-=ƒ11
x -2 (d) lim ƒ1x2 = 2 30.
lim
1x
5. 2x
c.x:0
d. -x +42sinx:
- x 21. lim sin
x:2x x:0 x 6. lim
(e) lim
31.
(a)
x:2
x:22.
-2 xlim
x:
1 + x:1
3
3
2x:12
2
x:0 x x
:
0
2
2x
+ 1x x + sin x
x 9 - 2x
+
x
2
sin x x + 7x + 12
x
x
lim+bƒ1x2 = 1
limdoes
ƒ1
lim ƒ1x2
(f ) (b)
lim lim 13.
lim
21.
22.
lim
14. lim
lim x:0
1e
sin
asinx:1
7. 23–26,
8.x lim ln (e)
11.
12.limlim
x : 1 x:1 +
find
the2xx2
limits.
x:0
x 1eInx Exercises
lim
sin
x2
ln
asin
b
7. x:
8.
x:
-1
x:2
2
2
x
+
1
x
2
x:0
x:p
(b)
2
-3
x:3 x + 2x - 15
x - 9x
(f) lim ƒ1x2 does not exist
x:0
x:p
(g) lim+ ƒ1x2 =
3 sin2x - 4 cos
(c)
In Exercises 23–26, find the limits. e2 - 1x
x : 0 lim ƒ1x
x
2 2 x x:1
23. lim x - 4
24. lim 3
x
4
ƒ
ƒ
x:1
3
2
x
1
lim
ƒ1x2
=
lim
ƒ1x2
(g)
:x0 5x + cos
x
(h) lim ƒ1x2(c)
exist
x + lim
1x : 0 log
2x
2+
41x + 222
ex - 15.
1x
x - 4 cos x x216.
lim
x:0 - 10.
1sin
lim
1x 3 -sin22
lim +2 x -x:0
x:c
2 x:9.
lim
lim
13.
14.
lim lim
lim
23.
24.
-2
x:
-2
x:a
x:a
x
+
2
x
+
2
ln12x2
1x
22
lim
9.
10.
1x +x9 - x2 (h)
+ 1lim ƒ1x2 exists for every
x:0x:
x:0 5 sin x + cos
log4-1
1x +x22+ 1
x 2
1-1,
12.
c in
ƒ1x2
(i) lim
exists
x:2
x:a
x:a
lim
25. lim
26.
x:c
x:c
x
+
1
In
Exe
(d)
2
x
:
p/2
x
:
27
sin 11–18,
x 1x (a)
ln12x2 In Exercises
3 xcannot
+ 9explain why2log
you
uselim
substitution
to
find33
x4 - 2 (i)
Exercises
31
and
ƒ1x2 exists
11,
32
for
every
c inand
.use
2
InIn
Exercises
34,
lim
lim
26.
Which
x
x:c
x
4
In25.
Exercises
11–18,
(a)
explain
why
you
cannot
use
substitution
to
find
lim
lim
1x
3
17.
18.
27–30,
use
the
given
graph
to
find
the
limits
or
to
explain
x:p/2 sin2 xIn Exercises
x:27
log
x
the
limit and (b) find316.
the limit
algebraically
if it exists.
Which
ofof,the
lim+ ƒ1x2
lim
(c)(e)
x:0
x:0 In Exercises
lim
lim
15.
33 and 34, (b)
use
ƒand
to statem
find
(a)
x2
why
the limits
do notalgebraically
exist.
xa:graph
0
x31
:0
the
limit
and
(b)
find
the
limit
if
it
exists.
x:
-2
x:
-2
x
+
2
x
+
2
In Exercises 27–30, use the given graph to find
2 the limits or to explain
2
x
+
7x
+
12
x
9
lim
ƒ1x2
lim
ƒ1x2
(b)
,
and
(c)
if
they
exist.
ƒ1x2(a)
= 11 lim
+ x2+1/xƒ
33.
31.
ƒ1x2
(a) lim
sin9lim
x x:0+
2 27. 11.
2y12.
why the limits do notxexist.
x:0
lim
32.
x:
-1
(g)
+
7x
+
12
x
x
:
2
2
2
1,ƒ1x2
In Exercises 19–22,
the-fact
that xlim
along
with
the
x: -3 usex12.
x:3=
1/x
2
9
x
+
2x
15
35.
Group
Activit
ƒ1x2
=
11
=
11
+
x2
33.
34.
lim
lim
11.
lim ƒ1x2
27.
(a) lim
x
(b) lim+ ƒ1x2 y
23
2 x:0
lim
ƒ1
(b)
17.
18.x:3lim
x:--31x 3 2x - 15 3
x:2
x:2
lim
g1x2
=
4.
Fin
x
9
x
+
3
2
2
x:0
x:0
x:0
ƒ1
lim
x
+
1
x
2x
+
x
2
35.
Group
Activity
Assume
that
x:4
(i)
x
limit
properties,
to
find
the
following
limits.
(b) lim+ ƒ1x23
2
ƒ1x2 3
(c) lim
x:4
lim
x
x:2
=
4. Find the (a)
limit.
2
lim 1g1x2
+ƒ1
x + 113. x :
x 3 -1 14.
2x 2 x:2
+limx - x:4
2lim xg1x2
lim
(c)
x:
-1 x + 1
2
x : 4 x:0 lim
lim
13.
14.
2
(c) lim ƒ1x2
sin x
sin x
sin 3x (a)
32.
(a)xƒ
x:2
2 limthe
(b)2 lim
x: -1
x:2 20.=xlim
x + lim
1use the fact
-, 2along
2 3 ƒ xxwith
21 that lim
1
In Exercises
19–22,
19.
-x:4
4 ƒ 1g1x2 + 22 (c) (d)
x
4
x:4 ƒ1
1
0
lim
g
1x2
lim
2
x:0
x:0 limx
x:0 x 2 16.
15.2xlim- x
x : 4 x:0 2
SECTIONSECTION
10.3 EXERCISES
10.3 EXERCISES
ƒ
x
ƒ
SECTION 10.3 EXERCISES
x
(b) lim+ ƒ1x2
x
x:1
7. lim 1e sin x2
8. lim ln asin b
x:0
x:p
2
(c) lim ƒ1x2
mbers with a gray background indicate problems that
29. (a) lim- ƒ1x2
y
x:1
2
x
:
3
x - 1
have designed
to
be
solved
without
a
calculator.
2
Example 2
Explain
why-the
be found
22 limit cannot10.
lim by direct
9. lim 1x
lim+ ƒ1x2 and find algebraically
(b) substitution
x:a
x:a x 2 + 1
4
x
:3
1–10, find the limit by direct
if thesubstitution
limit exists.if it exists.
In Exercises 31 and3 32, the graph of a functio
ƒ1x2
(c)
lim
2
In Exercises
11–18,
12 explain why you cannot use substitution to find
x:3
x 1x - 122
- 12(a)
2. lim 1x
Which of the statements about the function ar
1
x
:
3
the limit
and
(b)
find
the
limit
algebraically
if
it
exists.
x SECTION
a.
123
41
0ƒ1x2
3
3
lim
=
31.
(a)
1x - 2x + 32
4. lim 1x
2 - x + 52
x: -1 +
x2 - 9
x : -2 x + 7x + 12
lim
11. lim
12.
y = 0
(b) lim- ƒ1x2
ƒ1x2
x:3 x 2 +30.
lim-3 1x -x 242-2/39
1x + 5
6. x:
2x (a)
- 15xlim
x:0
: 1x : -2
x 3 + 1x
x 3 - 2x 2(b)
+ x lim
- 2 ƒ1x2
= 1
(c) lim- ƒ1x2
4
x
x:0
lim
lim
13.
14.
x : 1+
1e sin x2
ln
a
sin
b
8. lim
x:2
x + 763
12
x - 2
78.qxd 1/20/10 3:35 PM
xx:
: p-1Page
2 = lim+ ƒ1x2
(d) lim- ƒ1x2
(c) lim ƒ1x2
2
1
x:0
x:0
2
x
:
1
x
4
ƒ
ƒ
x
2x - 4
x
1
lim
lim
15.
16.
lim(a)
ƒ1x2
exists
(e)29.
(f)
1 2 ƒ1x2
3
0lim
Exercise
numbers with a gray background
indicate problems that
1x 2 - 22
10. lim
x: -2 x + 2
x: -2 x + 2
x:0
x
x:3
2
x : a xhave
the authors
+ 1designed to be solved without a calculator.
lim(b)
ƒ1x2
=ƒ1x2
1ƒ1x2
y =
31 and 32, the graph of a(g)
function
is given. (h) x
lim
xIn-Exercises
2
x:0
11–18, (a) explain why
you cannot
use
toby
find
x:3 +
lim substitution
lim 1x
- find
3substitution
17.
18.
In Exercises
1–10,
the
limit
direct
if
it
exists.
true=and
x:0
x:0 Which
ƒ1x2
0 which are false?
(i) limare
(j)
x 2 of the statements about the function
(b) find the limit algebraically if it exists. 2
x:1(c) lim ƒ1x2
x
12
y
x:3
lim
lim
x
1x
12
1x
12
1.
2.
lim + ƒ1x2SECTION
= 1 32.10.3
sinx:3
x 31. (a) x :
2
2
lim
ƒ1x2
=
1
(a)
More
on
Limits
763
x:
-1
x + 7x + 12
x = 1, along with-1the
In Exercises 19–22,
use9the fact that lim
x: -1 +
12. lim 2 3
x:0 x
3 (b) lim ƒ1x2 = 0
2
lim
1x
2x
+
32
lim
1x
x
+
52
3.
4.
3
x
:
3
3 exist
b. x:2x +to 2x
x - 9
(b) lim ƒ1x2 does not
x : 0limit properties,
find-the15following limits.x: -2
x:2
3
3
2
x + 1
- 2x + x - 2
lim ƒ1x2 = 1
(c) 2/3
lim
lim x1x
5. lim
6. lim
lim(a)
ƒ1x2
= 2ƒ1x2
(c)30.
x
sin x+ 5
sin-23x1x - 42x : 0 14.
x:1 - –1 1 2 3
x:2
x:
x:2
lim
19.
20.
1 x + 1
xlim
:2
x
2
SECTION 10.3x:0EXERCISES
(d)x lim- ƒ1x2 = lim+ ƒ1x2 (d) lim
x:0
x
lim= ƒ1x2
(b)ƒ1x2
2x 2x2 - x
2
x:0
2
x:1 - x:1 +
- 4x2ƒ
ƒ x2 sin
7. lim 1e
8. lim ln asin xb: 0
x - 4
x
x + sin x(e) 2 lim ƒ1x2 exists
x:p
lim sin
16. x:0
lim
ƒ1x2 1= 0
lim(c)
(e) (f)
umbers
problems that
29. (a) lim
21. xlim
22. lim
yƒ1x2
2 x + 2with a gray background
: -2 indicate
x : 0 ƒ1x2
0 lim=ƒ1x2
x
+
2
x:1x+:
x:1
x:3 x:0
x:0
x
2x
rs have designed to be solved without a calculator.
lim ƒ1x2
= 1
(g)
x 2 -(b)
1 xlim
lim lim
ƒ1x2ƒ1x2
(f) (h)
does=not1 exist
22
x
: 0+ ƒ1x2
x:1x4: 1
lim
1x
22
lim
9.
10.
In
Exercises
23–26,
find
the
limits.
x:3
lim
1x
3
18.
es 1–10, find the limit by direct
substitution
if it exists.
x:a x 2 + (i)
xx:a
:0
1 lim ƒ1x2 = 0
ƒ1x2
= 2 ƒ1x2
xx2
lim3lim
ƒ1x2
= lim
(g) (j)
lim
x :x1 ƒ1x2
e
1x
3 sin x -(c)4 cos
2
12
In Exercises
32,- the graph o
x:0x2+: 2 31 and
x:0
x:3
lim
m x 1x - 12
1x
12
2.
lim
lim cannot use substitution to find
23. x:3
24. you
In Exercises
11–18,
(a)
explain
why
sin
x
: -1
lim
ƒ1x2
=
1
32.
(a)
x every
x:0 5 sin x + cos x
+ 22 with the
ƒ1x2statements
(h) lim
exists for
in 1Which
of the
f
+
1, along
19–22, use the fact that lim x:0 log=431x
x:c0 1 2 3 4
find
the
limit algebraically if it exists.x : -1
x4.
: 0 and
x (b)
m 1x 3 - 2x + 32 the limit
lim
1x
x
+
52
1x + 9(b) lim ƒ1x2 does not exist (i)31.
:2
x: -2 ln12x2
lim(a)
ƒ1x2 lim
exists ƒ1x2
for every
= 1c in 11
2
ies, to find the following25.
limits.
lim x 2 2+ 7x2/3+ 12
26. lim
x9: 2
x:c
yx: -1 + y
x(a)
- lim
30.
ƒ1x2
lim
m 1x + 5
1x
42
6.
x:p/2
x:27
log
x
sin x
lim
11. x:
12. lim 23 (c) x:1
:2 sin x
-2
lim- ƒ1x2 = 2 In Exercises 33 (b)
and 34,
a graph
sin 3xx 2 - 9
limuseƒ1x2
= 0of ƒ to fi
x: -3
x:3 x + 2x x-:15
c.
lim 27–30, use
x:0 In 20.
Exercises
to2+explain
(b) orlim
ƒ1x2
x the given graph to find the limits
4
2x
x
:
0
lim
ƒ1x2
lim
ƒ1x2
(b)
,
and
(c)
if they exist
x
x:1 ƒ1x2 = 2
3
lim lndo
m2x1e -sinxx2
3 not exist.
why 8.
the x:p
limits
x:0 +
x:0ƒ1x2
x asin
+ 12 b
x 3 - (d)
2x 2xlim
+ 1x- - 2
=
1
(c)2 lim
2
:0
:
(c) lim ƒ1x2
x:0 -1/x
13. limx + sin x
14. lim
sin2 x
1
33. ƒ1x2 = 11 + x2
34. ƒ1x2
x: -1lim
x:2
(e)
y
x- +
1
x x:1
-lim
2 ƒ1x2 = 1
ƒ1x2
27. lim
(a)
22.
x1 2=3 limx ƒ1x
–1
ƒ1x2
(d)0 1lim
x : 1+
x 2 2x
- 1
2
3
x : 0 x:2
x
2
x:0
x:0 +that
m 1x - 22
10. lim 2x 2 - 4
35. Group Activity
Assume
x 2 (f
- )4 ƒlim ƒ1x2 does not exist
ƒ
(b)
:a
x:a xlim++ƒ1x2
x:1
23–26, find the limits. 15. limx:2 1
16. 3lim
lim g1x2
4.
Find
limit.
lim
ƒ1x2
exists
(e)
y =
=x:0
ƒ1x2
In Exercises 31 and 32, the graph of a function
is the
given.
x:4
x: -2 x + 2
2 -2 x +(g)2 lim+ ƒ1x2 = lim- ƒ1x2
es 11–18,
(a) explain why you(c)
cannot
lim use
ƒ1x2substitution to find x:
x
e - 1x
3 sin x - 4 cos x
Which of the xstatements
true
and+ƒ1x2
which
x:2
:0
x :the
0 function
lim(g)
1g1x2
22 =are1 false?
(a) are
(b)
lim
24. lim if it exists.
nd (b) find the limit algebraically
1 x - 2
x:4
x
x:0
x
:
0
log41x + 22
5
sin
x
+
cos
x
lim
1
1,
12
ƒ1x2
(h)
exists
for
every
c
in
.
y
17. lim 1x 2- 3
18. lim 31.2 (a) xlim
ƒ1x2 = 1
2
x
+
:
c
x:0
x:0
x + 7x + 12
x - 9
lim ƒ1x2 = 0
(i)g21x2
1 x 2 3 x: -1
0
lim
(c)
(d)
+ 9
m ln12x22
12. lim 1x
x:1
lim
ƒ1x2
11,
32
(i)
exists
for
every
c
in
.
x:4
x
2
lim
ƒ1x2
=
0
(b)
:
lim x + 2x - 15
m -3 2 x - 9
26. x:3
3
x
:
c
sin x
x:0
p/2 sin x
x : 27 log3 x
lim
ƒ1x2
=
1
32.
(a)
+
= (c)
1, along
In Exercises
19–22,
use
with
3 ƒ1x2
2 the fact that lim In
28. (a) xlim
36.ofGroup
Assume
that
x: -1
y Exercises
ƒ1x2
lim
33 and
34,the
ƒ to findActivity
(a)
,
x3 + 1
lim
ƒ1x2
=use
1 a graph
x:3 -- 2x + x - 2
x
x : 0x:0 27–30, use the given graph
find
the limits or to explain x:0 x
m
14. tolim
lim
g1x2
=
-3
.
Find
the
limit.
1
2
3
–1
lim
ƒ1x2
(b)
does
not
exis
(b) lim+ ƒ1x2, and (c) lim ƒ1x2 if they
exist.
: -1 x + 1
x:2 lim to
xfind
- 2the following limits.
ƒ1x2
(b)
x:a
limit properties,
x:2
ts do not exist.
x : 0(d) lim- ƒ1x2 =
x : lim
0 + ƒ1x2
x:3 +
2
x:0
x:0
3
lim(c)
1ƒ1x2
g1x22
(a) ƒ1x2
(b)
-x 4 ƒ
ƒ xsin
x 2ƒ1x2
- 4
= 2
= lim
11 ++ƒ1x2
x21/12x2
33.3xƒ1x2 = 11 + x21/x
34.
ƒ1x2
(c)
lim
2 sin
x:a
x
y
lim
m
lim
16.
lim
lim
ƒ1x2
exists
ƒ1x2
=
0
(e)
(f)
x:2
x:3
1
lim-2 x2 + 2
19.
20. lim
: 2x + 2
:x-2
x:
d. x:0
x:0
x:0
x
x:0 35.
x Group
ƒ1x2 ƒ1x2
= - 1=and
lim lim
Activity Assume that(d)
2x - x
2 (d)
0 1 2 3 4 (g)
lim ƒ1x2
lim
(c)
-+ 112
x13g1x2
:ƒ1x2
4x:1=
lim
lim
ƒ1x2
=
1
(h)
3 x -2 2
x : 2+
limx:0
g1x2 = 4. Find the limit. x:a
x:1
x + sin
x : 4x
mlim
1xƒ1x2
- 3
18. lim2 sin2 x
lim+=ƒ1x2
(e) ƒ1x2
lim x
lim
21.
22.
:0
x:0
lim
= 0+ 22
2 = 1
(i)
(j)(b)lim
- x:1
x:2
limƒ1x2
1g1x2
lim
xƒ1x2
(a)x:1
x:0
x:0
x
2x
x:2
1
x:4
x:4
sin x
lim ƒ1x2 does not exis
(f)
32. (a) lim + ƒ1x2 = 1
lim
= 1, along
es 19–22, use the factIn
that
with
the
xlimits.
x:1g1x2
Exercises
23–26,
find the
x: -1 2
1 2 3
x:0 x0
(c) lim g 1x2
(d) lim
lim ƒ1x2
(g)
lim
x : 4ƒ1x2 does not exist
x : 4 ƒ1x2
- 1 = lim- ƒ1x
3 sin(b)
x -x:2
4 cos x
rties, to find the following limits. ex - 1x
x:0 +
x:0
lim
lim
23.
24.
y
lim- ƒ1x2
ƒ1x2
=
2
lim
36.
Group
Activity
Assume
that
and
y
x:0
x:0
lim
ƒ1x2
=
2
(c)
log
1x
+
22
5
sin
x
+
cos
x
lim
ƒ1x2
(h)
exists
for eve
sin 3x
4
x : 3sin x
x : ax:c
m 2
20. lim
limx:2
g1x2 = - 3. Find the limit.
:0lim
x:0
ln12x2
x
3
x+: 9
a lim ƒ1x2 = 2
2x ƒ1x2
- x
1x(d)
(i) lim ƒ1x2
x : 3+
2 # exists for eve
x:1 lim
25. lim
26.
x:c
2
3 x + sin
lim
1ƒ1x2
+
g1x22
lim
1ƒ1x2
g1x22
(a)
(b)
2 x
x:p/2
x:27
sin
x
log(e)
ƒ1x2
3 x lim
x : a ƒ1x2 = 1
x:a
x
mxlim
22. lim21 sin x
+
:3
In
Exercises
33
and
34,
1
2 3use a graph
–1
x:1
:0
x:0
x
2x
ƒ1x2
x
In Exercises027–30,
limits or
to does
explain
1 2 3 4use the given graph to find the(f)
13g1x2
+not
12exist (b) lim
lim , and (c) lim ƒ1x2 if
limƒ1x2
(c) lim
(d)+ ƒ1x2
SECTION 10.3 EXERCISES
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