# The graph of a function can intersect a horizontal asymptote

## Full text

(1)

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)

(2)

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

(3)

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

### )

=0

6. , ,

,

7. , ,

, ,

8. , ,

9. If , then a calculator gives , , , , ,

, , , , , ,

, , .

It appears that .

(4)

f (x)= 1

 2/x

### )

x 0,10,000 0,0.2 lim

x  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]

(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.

(6)

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 .

(7)

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 ]

(8)

= 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

=lim

x 

9x2+x 3x

9x2+x +3x

9x2+x +3x

=lim

x 

9x2+x

2 3x

2

9x2+x +3x

=lim

x 

9x2+x

 9x2

9x2+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

= lim

x  

x+ x2+2x

x x2+2x

x 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

= 1

x x<0 x= x2

lim

x 

x2+ax x2+bx

=lim

x 

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

### ( )

/ x2

23.

24.

Note: In dividing numerator and denominator by , we used the fact that for , . 25.

(9)

=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

=lim

x  x

x 1

### )

= x  x 1  x 

lim

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

=lim

x  tan 1

x2

1 x2

t=x2

1 x2

### )

t  

x  x2  1 x2   lim

x  tan 1

x2

1 x2

### ) )

= lim

t   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,

(10)

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

= lim

x  

x2+x+1 +x

x2+x+1 x

x2+x+1 x

= lim

x  

x2+x+1

### ( )

 x2

x2+x+1 x

= lim

x  

(x+1)(1/x) x2+x+1 x

(1/x)= limx  

1+(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 .

(11)

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.

(12)

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.

(13)

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

### )

>0

x>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.

(14)

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=1

1 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

(15)

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.

(16)

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.

(17)

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.

(18)

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=750t

t 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)

(19)

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).

(20)

 =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).

(21)

 >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< lim

x  

1/x

### )

=0

M>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

M

lim

x  ex=

f

  ,a

lim

x   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

### )

=  lim

x  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

(22)

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 .

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