x f (x) 5
x f (x) 3
0 1 2
lim
x 2 f (x)=
lim
x 1
f (x)=
1. (a) As becomes large, the values of approach .
(b) As becomes large negative, the values of approach .
2. (a) The graph of a function can intersect a vertical asymptote in the sense that it can meet but not cross it.
The graph of a function can intersect a horizontal asymptote. It can even intersect its horizontal asymptote an infinite number of times.
(b) The graph of a function can have , , or horizontal asymptotes. Representative examples are shown.
No horizontal asymptote
One horizontal asymptote
Two horizontal asymptotes 3. (a)
(b)
lim
x 1+
f (x)=
lim
x f (x)=1
lim
x f (x)=2
x= 1 x=2 y=1 y=2
lim
x g(x)=2
lim
x g(x)= 2
lim
x 3g(x)=
lim
x 0g(x)=
lim
x 2+
g(x)=
x= 2 x=0 x=3 y= 2 y=2
f (0)=0 f (1)=1 lim
x f (x)=0 f
(c)
(d)
(e)
(f) Vertical: , ; Horizontal: , 4. (a)
(b)
(c)
(d)
(e)
(f) Vertical: , , ; Horizontal: ,
5. , , ,
is odd
lim
x 0+
f (x)= lim
x 0
f (x)= lim
x f (x)=1 lim
x f (x)=1
lim
x 2f (x)= lim
x f (x)= lim
x f (x)=0 lim
x 0+
f (x)= lim
x 0
f (x)=
lim
x 2f (x)= lim
x f (x)=3 lim
x f (x)= 3
f (x)=x2/2x f (0)=0 f (1)=0.5 f (2)=1 f (3)=1.125 f (4)=1
f (5)=0.78125 f (6)=0.5625 f (7)=0.3828125 f (8)=0.25 f (9)=0.158203125 f (10)=0.09765625 f (20) 0.00038147 f (50) 2.2204 10 12 f (100) 7.8886 10 27
lim
x
(
x2/2x)
=06. , ,
,
7. , ,
, ,
8. , ,
9. If , then a calculator gives , , , , ,
, , , , , ,
, , .
It appears that .
f (x)= 1
(
2/x)
x 0,10,000 0,0.2 limx f (x)=0.14
x f (x)
10,000 0.135308
100,000 0.135333 1,000,000 0.135335
lim
x f (x)=0.1353
lim
x
3x2 x+4 2x2+5x 8
=lim
x
(3x2 x+4)/x2 (2x2+5x 8)/x2
x2 x
= lim
x (3 1/x+4/x2) lim
x (2+5/x 8/x2)
= lim
x 3 lim
x (1/x)+lim
x (4/x2) lim
x 2+lim
x (5/x) lim
x (8/x2)
=
3 lim
x (1/x)+4lim
x (1/x2) 2+5lim
x (1/x) 8lim
x (1/x2)
= 3 0+4(0) 2+5(0) 8(0)
= 3 2
10. (a) From a graph of in a window of by , we estimate that (to two decimal places.)
(b)
From the table, we estimate that (to four decimal places.)
11.
[ divide both the numerator and denominator by
(the highest power of that appears in the denominator)]
[ Limit Law 5]
[ Limit Laws 1 and 2]
[ Limit Laws 7 and 3]
[Theorem 5 of Section 2.5]
lim
x
12x3 5x+2 1+4x2+3x3
= lim
x
12x3 5x+2 1+4x2+3x3
= lim
x
12 5/x2+2/x3 1/x3+4/x+3
x3
=
lim
x (12 5/x2+2/x3) lim
x (1/x3+4/x+3)
=
lim
x 12 lim
x (5/x2)+lim
x (2/x3) lim
x (1/x3)+lim
x (4/x)+lim
x 3
=
12 5lim
x (1/x2)+2lim
x (1/x3) lim
x (1/x3)+4lim
x (1/x)+3
= 12 5(0)+2(0) 0+4(0)+3
= 12
3 = 4 =2
lim
x
1
2x+3=lim
x
1/x (2x+3)/x=
lim
x (1/x) lim
x (2+3/x)=
lim
x (1/x) lim
x 2+3lim
x (1/x)= 0
2+3(0)=0 2 =0
lim
x
3x+5 x 4 =lim
x
(3x+5)/x (x 4)/x =lim
x
3+5/x 1 4/x =
lim
x 3+5lim
x
1 x lim
x 1 4lim
x
1 x
= 3+5(0) 1 4(0) =3 [ Limit Law 11]
[ divide by ]
[ Limit Law 5]
[ Limit Laws 1 and 2]
[ Limit Laws 7 and 3]
[Theorem 5 of Section 2.5]
13.
14.
lim
x
1 x x2
2x2 7 = lim
x
(1 x x2)/x2 (2x2 7)/x2
= lim
x (1/x2 1/x 1) lim
x (2 7/x2)
= lim
x (1/x2) lim
x (1/x) lim
x 1 lim
x 2 7 lim
x (1/x2)
= 0 0 1 2 7(0) = 1
2
lim
y
2 3y2 5y2+4y
=lim
y
(2 3y2)/y2 (5y2+4y)/y2
= lim
y (2/y2 3) lim
y (5+4/y) = 2lim
y (1/y2) lim
y 3 lim
y 5+4lim
y (1/y) = 2(0) 3 5+4(0)= 3
5
x3 x
lim
x
x3+5x 2x3 x2+4
=lim
x
x3+5x x3 2x3 x2+4
x3
=lim
x
1+ 5 x2 2 1
x+ 4 x3
= lim
x 1+ 5 x2 lim
x 2 1 x+ 4
x3
=
lim
x 1+5lim
x
1 x2 lim
x 2 lim
x
1
x+4lim
x
1 x3
= 1+5(0) 2 0+4(0)= 1
2
lim
t
t2+2
t3+t2 1= lim
t
t2+2
( )
/t3 t3+t2 1( )
/t3= lim
t
1/t+2/t3
1+1/t 1/t3= 0+0 1+0 0=0
u4 15.
16.
17. Divide both the numerator and denominator by (the highest power of that occurs in the denominator).
18.
19. First, multiply the factors in the denominator. Then divide both the numerator and denominator by .
lim
u
4u4+5
(u2 2)(2u2 1) =lim
u
4u4+5 2u4 5u2+2
=lim
u
4u4+5 u4 2u4 5u2+2
u4
=lim
u
4+ 5 u4 2 5
u2 + 2
u4
=
lim
u 4+ 5 u4 lim
u 2 5 u2
+ 2 u4
=
lim
u 4+5lim
u
1 u4 lim
u 2 5lim
u
1 u2
+2lim
u
1 u4
= 4+5(0) 2 5(0)+2(0)
= 4 2 =2
lim
x
x+2 9x2+1
=lim
x
(
x+2)
/x 9x2+1 / x2=lim
x
1+2/x 9+1/x2
= 1+0 9+0 = 1
3
lim
x
9x6 x x3+1
=lim
x
9x6 x /x3 (x3+1)/x3
= lim
x (9x6 x)/x6 lim
x (1+1/x3)
x3= x6 x>0
= lim
x 9 1/x5 lim
x 1+lim
x (1/x3)
=
lim
x 9 lim
x (1/x5) 1+0
= 9 0=3
lim
x
9x6 x x3+1
= lim
x
9x6 x /x3 (x3+1)/x3
= lim
x (9x6 x)/x6 lim
x (1+1/x3)
x3= x6 x<0 20.
21.
[ since for ]
22.
[ since for ]
= lim
x 9 1/x5 lim
x 1+ lim
x (1/x3)
=
lim
x 9 lim
x (1/x5) 1+0
= 9 0= 3
lim
x
(
9x2+x 3x)
=limx
9x2+x 3x
( ) (
9x2+x +3x)
9x2+x +3x
=lim
x
9x2+x
( )
2 3x( )
29x2+x +3x
=lim
x
9x2+x
( )
9x29x2+x +3x
=lim
x
x 9x2+x +3x
1/x 1/x
=lim
x
x/x
9x2/x2+x/x2+3x/x
=lim
x
1
9+1/x +3= 1
9 +3= 1 3+3 =1
6
lim
x
(
x+ x2+2x)
= limx
(
x+ x2+2x)
x x2+2xx x2+2x
= lim
x
x2 x
(
2+2x)
x x2+2x
= lim
x
2x x x2+2x
= lim
x
2
1+ 1+2/x = 2
1+ 1+2 0
( )
= 1x x<0 x= x2
lim
x
(
x2+ax x2+bx)
=limx
x2+ax x2+bx
( ) (
x2+ax + x2+bx)
x2+ax + x2+bx
=lim
x
(x2+ax) (x2+bx) x2+ax + x2+bx
=lim
x
[(a b)x]/x x2+ax + x2+bx
( )
/ x223.
24.
Note: In dividing numerator and denominator by , we used the fact that for , . 25.
=lim
x
a b
1+a/x + 1+b/x = a b
1+0 + 1+0 = a b 2 lim
x cos x x cos x
1 1
x x lim
x x =
3 x x lim
x
3 x = lim
x
(
x x)
=limx x
(
x 1)
= x x 1 xlim
x
x3 2x+3 5 2x2
=lim
x
(x3 2x+3)/x2 (5 2x2)/x2
x
=lim
x
x 2/x+3/x2
5/x2 2 = x 2/x+3/x2 5/x2 2 2 x . lim
x (x4+x5)= lim
x x5( 1
x+1)= x5 1/x+1 1 x
lim
x tan 1
(
x2 x4)
=limx tan 1
(
x2(
1 x2) )
t=x2(
1 x2)
tx x2 1 x2 lim
x tan 1
(
x2(
1 x2) )
= limt tan 1t= 2
lim
x
x+x3+x5 1 x2+x4
=lim
x
(x+x3+x5)/x4 (1 x2+x4)/x4
x
=lim
x
1/x3+1/x+x 1/x4 1/x2+1
=
(1/x3+1/x+x) (1/x4 1/x2+1) 1 x t=tan x x ( /2)+ t
26. does not exist because as increases does not approach any one value, but oscillates between and .
27. is large when is large, so .
28. is large negative when is large negative, so .
29. since and as .
30. [divide by the highest power of in the denominator]
because and as
31. because and as .
32. . If we let , we know that as
, since and . So .
33.
[ divide by the highest power of in the denominator ]
because and as .
34. If we let , then as , . Thus,
lim
x ( /2)+
etan x= lim
t et=0
f (x)= x2+x+1 +x lim
x f (x) 0.5
x f (x)
10,000 0.4999625
100,000 0.4999962
1,000,000 0.4999996
0.5
lim
x
(
x2+x+1 +x)
= limx
(
x2+x+1 +x)
x2+x+1 xx2+x+1 x
= lim
x
x2+x+1
( )
x2x2+x+1 x
= lim
x
(x+1)(1/x) x2+x+1 x
( )
(1/x)= limx1+(1/x)
1+(1/x)+ 1/x2
( )
1= 1+0
1+0+0 1= 1 2
x<0 x2= x = x x x<0
1
x x2+x+1 = 1 x2
x2+x+1 = 1+(1/x)+ 1/x
(
2)
.
35. (a)
From the graph of , we estimate the value of to be .
(b)
From the table, we estimate the limit to be . (c)
Note that for , we have , so when we divide the radical by , with , we get .
f (x)= 3x2+8x+6 3x2+3x+1 lim
x f (x) 1.4
x f (x)
10,000 1.44339
100,000 1.44338 1,000,000 1.44338
1.4434
lim
x f (x) =lim
x
3x2+8x+6 3x2+3x+1
( ) (
3x2+8x+6 + 3x2+3x+1)
3x2+8x+6 + 3x2+3x+1
=lim
x
3x2+8x+6
( )
3x(
2+3x+1)
3x2+8x+6 + 3x2+3x+1
=lim
x
(5x+5)(1/x) 3x2+8x+6 + 3x2+3x+1
( )
(1/x)=lim
x
5+5/x
3+8/x+6/x2+ 3+3/x+1/x2
= 5
3 + 3 = 5
2 3 = 5 3
6 1.443376
lim
x
x
x+4 = lim
x
1
1+4/x = 1
1+0 =1 y=1 lim
x 4
x x+4 = lim
x 4+
x
x+4 = x= 4
From the graph of , we estimate (to one decimal place) the value of to be .
(b)
From the table, we estimate (to four decimal places) the limit to be . (c)
37. , so is a horizontal asymptote. and
, so is a vertical asymptote. The graph confirms these calculations.
x2 1 0 x 1 y<0 1<x<1 y>0 x< 1 x>1 lim
x 1
x2+4
x2 1 = lim
x 1+
x2+4
x2 1= lim
x 1
x2+4
x2 1 = lim
x 1+
x2+4
x2 1 = x=1 x= 1 lim
x
x2+4 x2 1 = lim
x
1+4/x2 1 1/x2
=1+0
1 0 =1 y=1
lim
x
x3
x2+3x 10 = lim
x
x
1+(3/x) 10/x
(
2)
=lim
x 2+
x3
x2+3x 10 =lim
x 2+
x3
(x+5)(x 2)= x3
(x+5)(x 2) >0 x>2 lim
x 2
x3
x2+3x 10 = lim
x 5
x3
x2+3x 10 = lim
x 5+
x3
x2+3x 10 = x=2 x= 5
38. Since as and for and for and , we have
, , , and , so and are
vertical asymptotes. Also , so is a horizontal asymptote.
The graph confirms these calculations.
39. , so there is no horizontal asymptote.
, since for . Similarly,
, , and , so and are vertical asymptotes. The
graph confirms these calculations.
lim
x
x3+1 x3+x
= lim
x
1+1/x3 1+1/x2
=1 y=1 y= x3+1
x3+x
= x3+1 x x
(
2+1)
>0x>0 y<0 1<x<0 lim
x 0+
x3+1 x3+x
= lim
x 0
x3+1 x3+x
= x=0
lim
x
x
4
x4+1
1/x
1/4 x4
=lim
x
1
4 1+ 1 x4
= 1
4 1+0
=1
lim
x
x
4
x4+1
1/x
1/4 x4
= lim
x
1
4 1+ 1 x4
= 1
4 1+0
= 1 y= 1
lim
x
x 9 4x2+3x+2
=lim
x
1 9/x
4+ 3/x
( )
+ 2/x(
2)
=1 0
4+0+0 =1 2
x2= x = x x<0 x
x2 lim
x
x 9 4x2+3x+2
= lim
x
1+9/x
4+ 3/x
( )
+ 2/x(
2)
=1+0
4+0+0 = 1 2 y= 1
2 4x2+3x+2 x
, so is a horizontal asymptote. Since for
and for , and , so is a vertical asymptote.
41. and
, so are horizontal asymptotes.
There is no vertical asymptote.
42. .
Using the fact that for , we divide the numerator by and the denominator by .
Thus, .
The horizontal asymptotes are . The polynomial is positive for all , so the denominator never approaches zero, and thus there is no vertical asymptote.
lim
x f (x)=0 <
lim
x 0 f (x)= x2 x
x=0 x=0
lim
x 3
f (x)= lim
x 3+
f (x)= x=3
(
x 3)
f (2)=0 2 x
(
x 2)
f (x)= 2 x x2(x 3)
x=1 x=3
x 1
( ) (
x 3)
y=11 f (x)= x2
(x 1)(x 3)
y= f (x)=x2(x 2)(1 x) y f (0)=0 x y=0 x=0 1
2 x2 x 0
x 1 2 lim
x x2(x 2)(1 x)= 43. Let’s look for a rational function.
(1) degree of numerator degree of denominator
(2) there is a factor of in the denominator (not just , since that would produce a sign change at ), and the function is negative near .
(3) and vertical asymptote at ; there is a factor of in the denominator.
(4) is an intercept; there is at least one factor of in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left and right hand limits gives us as one possibility.
44. Since the function has vertical asymptotes and , the denominator of the rational function we are looking for must have factors and . Because the horizontal asymptote is , the degree of the numerator must equal the degree of the denominator, and the ratio of the leading
coefficients must be . One possibility is .
45. . The intercept is , and the intercepts occur when , , and . Notice that, since is always positive, the graph does not cross the axis at , but does cross the axis at and . , since the first two factors are large positive and
lim
x x2(x 2)(1 x)= x
y=(2+x)3(1 x)(3 x) x lim
x f (x)= x lim
x f (x)= y
f (0)=(2)3(1)(3)=24 x f (x)=0 x= 2 1 3
x
y= f (x)=(x+4)5(x 3)4 y f (0)=45( 3)4=82 944 x
y=0 x= 4 3 x 3 (x 3)4
x 4 lim
x (x+4)5(x 3)4=
x lim
x (x+4)5(x 3)4= x
because the first and third factors are large positive and the second large negative as .
46. . As , the first factor is large positive, and the second and third factors are large negative. Therefore, . As , the first factor is large negative, and the second and third factors are large positive. Therefore, . Now the intercept is
and the intercepts are the solutions to , and , and the graph crosses the axis at all of these points.
47. . The intercept is , . The intercepts occur when
, . Notice that the graph does not cross the axis at because is always positive, but does cross the axis at . since both factors are large positive when is large positive. since the first factor is large negative and the second factor is large positive when is large negative.
y=(1 x)(x 3)2(x 5)2 x
lim
x (x)= x
lim
x (x)= y f (0)=(1)( 3)2( 5)2=225 x
f (x)=0 x=1 3 5 f (x) x=3 x=5
(x 3)2 (x 5)2 x x=3 x=5
x x=1
1 sin x 1 x, 1
x sin x x 1
x x>0 x 1/x 0 1/x 0 (sin x)/x 0 lim
x
sin x x =0
y=0 y=(sin x)/x
sin x=0 x= n n
lim
x P(x)=lim
x Q(x)= lim
x P(x)= lim
x Q(x)=
P Q
48. . As , the first factor approaches while the second and third factors approach . Therefore, . As , the factors all approach . Therefore,
. Now the intercept is and the intercepts are the solutions to , , and . Notice that does not change sign at or because the factors and are always positive, so the graph does not cross the axis at or , but does cross the axis at .
49. (a) Since for all for . As , and , so by
the Squeeze Theorem, . Thus, .
(b) From part (a), the horizontal asymptote is . The function crosses the horizontal asymptote whenever ; that is, at for every integer . Thus, the graph crosses the asymptote an infinite number of times.
50. (a) In both viewing rectangles, and . In the
larger viewing rectangle, and become less distinguishable.
lim
x
P(x) Q(x)=lim
x
3x5 5x3+2x 3x5
=lim
x 1 5 3 1
x2 +2
3 1 x4
=1 5
3(0)+2 3(0)=1
P Q
x Q(x)
deg P<deg Q 0 lim
x P(x)/Q(x) =0
deg P>deg Q
lim
x P(x)/Q(x) = P Q
n=0 n>0 (n odd) n>0 (n even) n<0 (n odd) n<0 (n even)
lim
x 0+
xn=
1 0
if n=0 if n>0 if n<0
{
lim
x 0
xn=
1
0
if n=0 if n>0
if n<0, n odd if n<0, n even
{
lim
x xn=
1 0
if n=0 if n>0 if n<0
{
lim
x xn=
1
0
if n=0
if n>0, n odd if n>0, n even if n<0
{
(b)
and have the same end behavior.
51. Divide the numerator and the denominator by the highest power of in .
(a) If , then the numerator but the denominator doesn’t. So .
(b) If , then the numerator but the denominator doesn’t, so (depending on the ratio of the leading coefficients of and ).
52.
(i) (ii) (iii) (iv) (v)
(a)
(b)
(c)
(d)
53.
lim
x
4x 1 x =lim
x 4 1
x =4 lim
x
4x2+3x x2
=lim
x 4+ 3 x =4 lim
x f (x)=4
t 25t 30
5000+25t
( )
25t 30=750tt C(t)= 750t
5000+25t = 30t 200+t
g L lim
t C(t)=lim
t
30t
200+t =lim
t
30t/t
200/t+t/t = 30 0+1 =30
lim
t v(t)=lim
t v* 1 e gt/v
(
*)
=v*(1 0)=v*v(t)=1 e 9.8t v(t)=0.99v* v(t)=0.99 t 0.47
y=e x/10 y=0.1 x
1 23.03 x>x
1 e x/10<0.1 e x/10<0.1
x> 10ln 1
10 = 10ln 10 1=10ln 10
, and . Therefore, by the Squeeze
Theorem, .
54. (a) After minutes, liters of brine with g of salt per liter has been pumped into the tank, so it contains liters of water and grams of salt. Therefore, the salt
concentration at time will be .
(b) . So the salt concentration approaches that of the
brine being pumped into the tank.
55. (a)
(b) We graph and , or in this case, . Using an intersect feature or zooming in on the point of intersection, we find that s.
56. (a) and intersect at .
If , then .
(b)
6x2+5x 3
2x2 1 3 <0.2 2.8<6x2+5x 3 2x2 1 <3.2
y= 6x2+5x 3
2x2 1 y=2.8 y=3.2
x>12.8 N=13
x N
=0.5 N x N 4x2+1
x+1 2 <0.5 1.5< 4x2+1
x+1 <2.5
x 3 N=3 =0.1
1.9< 4x2+1
x+1 <2.1 x 19 N=19
57. . So we graph the three parts of this inequality on
the same screen, and find that the curve seems to lie between the lines and whenever . So we can choose (or any larger number) so that the inequality holds whenever .
58. For , we must find such that whenever , we have
. We graph the three parts of this inequality on the same screen, and find that it holds whenever . So we choose (or any larger number). For , we must have
, and the graphs show that this holds whenever . So we choose (or any larger number).
=0.5 N 4x2+1
x+1 ( 2) <0.5 2.5< 4x2+1
x+1 < 1.5 x N
x 6. N= 6
=0.1 2.1< 4x2+1
x+1 < 1.9 x N.
x 22 N= 22
N 2x+1
x+1 >100 x N.
x 2500 N=2500
59. For , we need to find such that
whenever . We graph the three parts of this inequality on the same screen, and see that the inequality holds for So we choose (or any smaller number).
For , we need whenever From the graph, it seems that this inequality holds for . So we choose (or any smaller number).
60. We need such that whenever From the graph, we see that this inequality holds for . So we choose (or any larger number).
>0 1/x2< x2>1/ x>1/ N=1/ x>N x> 1
1
x2
0 = 1 x2
< lim
x
1 x2
=0
1/ x <0.0001 x >1/0.0001=104 x>108
>0 1/ x < x >1/ x>1/ 2 N=1/ 2
x>N x> 1
2 1
x 0 = 1
x < lim
x
1 x =0 x<0 1/x 0 = 1/x >0 1/x< x< 1/
N= 1/ x<N
(
1/x)
0 = 1/x< limx
(
1/x)
=0M>0 N>0 x>N x3>M x3>M x>3 M N=3 M x>N=3 M x3>M lim
x x3=
M>0 N>0 x>N ex>M ex>M x>ln M
N=max 1,ln M
( )
N>0 x>N=max 1,ln M( )
ex>max e,M( )
Mlim
x ex=
f
(
,a)
limx f (x)=
M N f (x)<M
x<N lim
x
(
1+x3)
=M N x<N 1+x3<M 1+x3<M x3<M 1
x<3 M 1 N=3 M 1 x<N 1+x3<M lim
x
(
1+x3)
= limx f (x)=L >0 N
f (x) L < x>N t=1/x x>N 0<1/x<1/N 0<t<1/N >0
>0 1/N f (1/t) L < 0<t<
(b) If is given, then . Let .
Then , so .
62. (a)
(b) If is given, then . Let .
Then , so .
63. For , . If is given, then .
Take . Then , so .
64. Given , we need such that . Now , so take . Then
, so .
65. Given , we need such that . Now , so take
. (This ensures that .) Then , so
.
66. Definition Let be a function defined on some interval . Then means that for every negative number there is a corresponding negative number such that
whenever . Now we use the definition to prove that . Given a negative number , we need a negative number such that . Now
. Thus, we take and find that . This proves that .
67. Suppose that . Then for every there is a corresponding positive number such
that whenever . If , then . Thus, for every
there is a corresponding (namely ) such that whenever . This proves that
lim
t 0+
f (1/t)=L=lim
x f (x) lim
x f (x)=L >0 N
f (x) L < x<N t=1/x x<N 1/N<1/x<0 1/N<t<0
>0 >0 1/N f (1/t) L < <t<0
lim
t 0
f (1/t)=L= lim
x f (x) .
Now suppose that . Then for every there is a corresponding negative number
such that whenever . If , then . Thus, for every
there is a corresponding (namely ) such that whenever . This
proves that .