Section 2.4 The Precise Deﬁnition of a Limit

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Section 2.4 The Precise Definition of a Limit

SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 91

Solution 2: We add a few lines to the diagram, as shown. Note that

∠ = 90^◦(subtended by diameter  ). So∠ = 90^◦=∠

(subtended by diameter  ). It follows that∠ = ∠. Also

∠ = 90^◦− ∠  = ∠ . Since 4 is isosceles, so is 4 , implying that  =  . As the circle ²shrinks, the point  plainly approaches the origin, so the point  must approach a point twice as far from the origin as  , that is, the point (4 0), as above.

2.4 The Precise Definition of a Limit

1.If |() − 1|  02, then −02  () − 1  02 ⇒ 08  ()  12. From the graph, we see that the last inequality is true if 07    11, so we can choose  = min {1 − 07 11 − 1} = min {03 01} = 01 (or any smaller positive number).

2.If |() − 2|  05, then −05  () − 2  05 ⇒ 15  ()  25. From the graph, we see that the last inequality is true if 26    38, so we can take  = min {3 − 26 38 − 3} = min {04 08} = 04 (or any smaller positive number).

Note that  6= 3.

3.The leftmost question mark is the solution of √ = 16 and the rightmost, √ = 24. So the values are 16² = 256and 24²= 576. On the left side, we need | − 4|  |256 − 4| = 144. On the right side, we need | − 4|  |576 − 4| = 176.

To satisfy both conditions, we need the more restrictive condition to hold — namely, | − 4|  144. Thus, we can choose

 = 144, or any smaller positive number.

4.The leftmost question mark is the positive solution of ²=¹₂, that is,  = √¹

2, and the rightmost question mark is the positive solution of ²= ³₂, that is,  =

3

2. On the left side, we need | − 1| ^√¹2 − 1 ≈ 0292 (rounding down to be safe). On the right side, we need | − 1| 

3 2− 1



 ≈ 0224. The more restrictive of these two conditions must apply, so we choose

 = 0224(or any smaller positive number).

5. From the graph, we ﬁnd that  = tan  = 08 when  ≈ 0675, so



4 − ¹≈ 0675 ⇒ ¹≈^4 − 0675 ≈ 01106. Also,  = tan  = 12 when  ≈ 0876, so^4 + 2≈ 0876 ⇒ ²= 0876 −^4 ≈ 00906.

Thus, we choose  = 00906 (or any smaller positive number) since this is the smaller of 1and 2.

6. From the graph, we ﬁnd that  = 2(²+ 4) = 03when  = ²₃, so

1 − ¹=²₃ ⇒ ¹=¹₃. Also,  = 2(²+ 4) = 05when  = 2, so 1 + 2= 2 ⇒ ²= 1. Thus, we choose  = ¹₃ (or any smaller positive number) since this is the smaller of 1and 2.

(c)  is the radius, () is the area,  is the target radius given in part (a),  is the target area (1000 cm²),  is the magnitude of the error tolerance in the area (5 cm²), and  is the tolerance in the radius given in part (b).

12. (a)  = 01²+ 2155 + 20and  = 200 ⇒

01²+ 2155 + 20 = 200 ⇒ [by the quadratic formula or from the graph]  ≈ 330 watts (  0)

(b) From the graph, 199 ≤  ≤ 201 ⇒ 3289    3311.

(c)  is the input power, () is the temperature,  is the target input power given in part (a),  is the target temperature (200),

is the tolerance in the temperature (1), and  is the tolerance in the power input in watts indicated in part (b) (011 watts).

13. (a) |5 − 15| = 5 | − 3|  01 ⇔ | − 3|  01

5 = 002, so  = 002.

(b) |5 − 15| = 5 | − 3|  001 ⇔ | − 3| 001

5 = 0002, so  = 0002.

14. |(5 − 7) − 3| = |5 − 10| = |5( − 2)| = 5 | − 2|. We must have |() − |  , so 5 | − 2|   ⇔

| − 2|  5. Thus, choose  = 5. For  = 01,  = 002; for  = 005,  = 001; for  = 001,  = 0002.

15. Given   0, we need   0 such that if 0  | − 1|  , then

|(2 + 3) − 5|  . But |(2 + 3) − 5|   ⇔

|2 − 2|   ⇔ 2 | − 1|   ⇔ | − 1|  2.

So if we choose  = 2, then 0  | − 1|   ⇒

|(2 + 3) − 5|  . Thus, lim

→1(2 + 3) = 5by the deﬁnition of a limit.

|(2 − 5) − 3|  . But |(2 − 5) − 3|   ⇔ |2 − 8|   ⇔

|2| | − 4|   ⇔ | − 4|  2. So if we choose  = 2, then 0  | − 4|   ⇒ |(2 − 5) − 3|  . Thus, lim_

→4(2 − 5) = 3 by the deﬁnition of a limit.

|(2 + 3) − 5|  . But |(2 + 3) − 5|   ⇔

|2 − 2|   ⇔ 2 | − 1|   ⇔ | − 1|  2.

So if we choose  = 2, then 0  | − 1|   ⇒

|(2 + 3) − 5|  . Thus, lim

→1(2 + 3) = 5by the deﬁnition of a limit.

24. Given   0, we need   0 such that if 0  | − |  , then | − |  . But | − | = 0, so this will be true no matter what  we pick.

25. Given   0, we need   0 such that if 0  | − 0|  , then²− 0

   ⇔ ²  ⇔ || √. Take  = √.

Then 0  | − 0|   ⇒ ²− 0  . Thus, lim

→0²= 0by the deﬁnition of a limit.

26. Given   0, we need   0 such that if 0  | − 0|  , then³− 0

   ⇔ ||³  ⇔ || √³. Take  =√³.

Then 0  | − 0|   ⇒ ³− 0  ³= . Thus, lim

→0³= 0by the deﬁnition of a limit.

27. Given   0, we need   0 such that if 0  | − 0|  , then

|| − 0  . But

||

 = ||. So this is true if we pick  = .

Thus, lim

→0|| = 0 by the deﬁnition of a limit.

28. Given   0, we need   0 such that if 0   − (−6)  , then√⁸

6 +  − 0  . But√⁸

6 +  − 0

   ⇔

√8

6 +    ⇔ 6 +   ⁸ ⇔  − (−6)  ⁸. So if we choose  = ⁸, then 0   − (−6)   ⇒

√⁸

6 +  − 0  . Thus, lim

→−6⁺

√8

6 +  = 0by the deﬁnition of a right-hand limit.

29. Given   0, we need   0 such that if 0  | − 2|  , then²− 4 + 5

− 1

   ⇔ ²− 4 + 4

   ⇔

( − 2)²

  . So take  = √. Then 0  | − 2|   ⇔ | − 2| √

 ⇔ 

( − 2)²  . Thus,

lim→2

²− 4 + 5

= 1by the deﬁnition of a limit.

30. Given   0, we need   0 such that if 0  | − 2|  , then(²+ 2 − 7) − 1  . But(²+ 2 − 7) − 1

   ⇔

²+ 2 − 8

   ⇔ | + 4| | − 2|  . Thus our goal is to make | − 2| small enough so that its product with | + 4|

is less than . Suppose we ﬁrst require that | − 2|  1. Then −1   − 2  1 ⇒ 1    3 ⇒ 5   + 4  7 ⇒

| + 4|  7, and this gives us 7 | − 2|   ⇒ | − 2|  7. Choose  = min {1 7}. Then if 0  | − 2|  , we have | − 2|  7 and | + 4|  7, so(²+ 2 − 7) − 1

 = |( + 4)( − 2)| = | + 4| | − 2|  7(7) = , as desired. Thus, lim

→2(²+ 2 − 7) = 1 by the deﬁnition of a limit.

31. Given   0, we need   0 such that if 0  | − (−2)|  , then²− 1

− 3   or upon simplifying we need

²− 4

   whenever 0  | + 2|  . Notice that if | + 2|  1, then −1   + 2  1 ⇒ −5   − 2  −3 ⇒

| − 2|  5. So take  = min {5 1}. Then 0  | + 2|   ⇒ | − 2|  5 and | + 2|  5, so

²− 1

− 3

 = |( + 2)( − 2)| = | + 2| | − 2|  (5)(5) = . Thus, by the deﬁnition of a limit, lim

→−2(²− 1) = 3.

32. Given   0, we need   0 such that if 0  | − 2|  , then³− 8  . Now³− 8 =

( − 2)

²+ 2 + 4.

If | − 2|  1, that is, 1    3, then ²+ 2 + 4  3²+ 2(3) + 4 = 19and so

³− 8

 = | − 2|

²+ 2 + 4

 19 | − 2|. So if we take  = min 1₁₉^

, then 0  | − 2|   ⇒

³− 8

 = | − 2|

²+ 2 + 4

₁₉^ · 19 = . Thus, by the deﬁnition of a limit, lim_

→2³= 8.

24. Given   0, we need   0 such that if 0  | − |  , then | − |  . But | − | = 0, so this will be true no matter what  we pick.

25. Given   0, we need   0 such that if 0  | − 0|  , then²− 0

   ⇔ ²  ⇔ || √. Take  = √.

Then 0  | − 0|   ⇒ ²− 0  . Thus, lim

→0²= 0by the deﬁnition of a limit.

26. Given   0, we need   0 such that if 0  | − 0|  , then³− 0

   ⇔ ||³  ⇔ || √³. Take  =√³.

Then 0  | − 0|   ⇒ ³− 0  ³= . Thus, lim

→0³= 0by the deﬁnition of a limit.

27. Given   0, we need   0 such that if 0  | − 0|  , then

|| − 0  . But

||

 = ||. So this is true if we pick  = .

Thus, lim

→0|| = 0 by the deﬁnition of a limit.

28. Given   0, we need   0 such that if 0   − (−6)  , then√⁸

6 +  − 0  . But√⁸

6 +  − 0

   ⇔

√8

6 +    ⇔ 6 +   ⁸ ⇔  − (−6)  ⁸. So if we choose  = ⁸, then 0   − (−6)   ⇒

√⁸

6 +  − 0  . Thus, lim

→−6⁺

√8

6 +  = 0by the deﬁnition of a right-hand limit.

29. Given   0, we need   0 such that if 0  | − 2|  , then²− 4 + 5

− 1

   ⇔ ²− 4 + 4

   ⇔

( − 2)²

  . So take  = √. Then 0  | − 2|   ⇔ | − 2| √

 ⇔ 

( − 2)²  . Thus,

lim→2

²− 4 + 5

= 1by the deﬁnition of a limit.

30. Given   0, we need   0 such that if 0  | − 2|  , then(²+ 2 − 7) − 1  . But(²+ 2 − 7) − 1

   ⇔

²+ 2 − 8

   ⇔ | + 4| | − 2|  . Thus our goal is to make | − 2| small enough so that its product with | + 4|

is less than . Suppose we ﬁrst require that | − 2|  1. Then −1   − 2  1 ⇒ 1    3 ⇒ 5   + 4  7 ⇒

| + 4|  7, and this gives us 7 | − 2|   ⇒ | − 2|  7. Choose  = min {1 7}. Then if 0  | − 2|  , we have | − 2|  7 and | + 4|  7, so(²+ 2 − 7) − 1

 = |( + 4)( − 2)| = | + 4| | − 2|  7(7) = , as desired. Thus, lim

→2(²+ 2 − 7) = 1 by the deﬁnition of a limit.

31. Given   0, we need   0 such that if 0  | − (−2)|  , then²− 1

− 3   or upon simplifying we need

²− 4

   whenever 0  | + 2|  . Notice that if | + 2|  1, then −1   + 2  1 ⇒ −5   − 2  −3 ⇒

| − 2|  5. So take  = min {5 1}. Then 0  | + 2|   ⇒ | − 2|  5 and | + 2|  5, so

²− 1

− 3

 = |( + 2)( − 2)| = | + 2| | − 2|  (5)(5) = . Thus, by the deﬁnition of a limit, lim_

→−2(²− 1) = 3.

32. Given   0, we need   0 such that if 0  | − 2|  , then³− 8  . Now³− 8 =

( − 2)

²+ 2 + 4.

If | − 2|  1, that is, 1    3, then ²+ 2 + 4  3²+ 2(3) + 4 = 19and so

³− 8

 = | − 2|

²+ 2 + 4

 19 | − 2|. So if we take  = min 1₁₉^

, then 0  | − 2|   ⇒

³− 8

 = | − 2|

²+ 2 + 4

₁₉^ · 19 = . Thus, by the deﬁnition of a limit, lim

→2³= 8.

98 ¤ CHAPTER 2 LIMITS AND DERIVATIVES

43. Given   0 we need   0 so that ln    whenever 0    ; that is,  = ^{ln } ^ whenever 0    . This suggests that we take  = ^. If 0    ^, then ln   ln ^ = . By the deﬁnition of a limit, lim

→0⁺ln  = −∞.

44. (a) Let  be given. Since lim

→ () = ∞, there exists ¹ 0such that 0  | − |  ¹ ⇒ ()   + 1 − . Since

lim→() = , there exists 2  0such that 0  | − |  ² ⇒ |() − |  1 ⇒ ()   − 1. Let  be the smaller of 1and 2. Then 0  | − |   ⇒ () + ()  ( + 1 − ) + ( − 1) = . Thus,

lim→[ () + ()] = ∞.

(b) Let   0 be given. Since lim

→() =   0, there exists 1  0such that 0  | − |  ¹ ⇒

|() − |  2 ⇒ ()  2. Since lim

→ () = ∞, there exists ² 0such that 0  | − |  ² ⇒

 ()  2. Let  = min {¹ 2}. Then 0  | − |   ⇒ () () 2



2 = , so lim

→ () () = ∞.

(c) Let   0 be given. Since lim

→() =   0, there exists 1 0such that 0  | − |  ¹ ⇒

|() − |  −2 ⇒ ()  2. Since lim_

→ () = ∞, there exists ² 0such that 0  | − |  ² ⇒

 ()  2. (Note that   0 and   0 ⇒ 2  0.) Let  = min {¹ 2}. Then 0  | − |   ⇒

 ()  2 ⇒ () () 2

 · 

2 = , so lim

→ () () = −∞.

2.5 Continuity

1. From Deﬁnition 1, lim

→4 () =  (4).

2. The graph of  has no hole, jump, or vertical asymptote.

3. (a)  is discontinuous at −4 since (−4) is not deﬁned and at −2, 2, and 4 since the limit does not exist (the left and right limits are not the same).

(b)  is continuous from the left at −2 since lim

→−2⁻ () =  (−2).  is continuous from the right at 2 and 4 since lim

→2⁺ () =  (2)and lim

→4⁺ () =  (4). It is continuous from neither side at −4 since (−4) is undeﬁned.

4. From the graph of , we see that  is continuous on the intervals [−3 −2), (−2 −1), (−1 0], (0 1), and (1 3].

5. The graph of  = () must have a discontinuity at

 = 2and must show that lim

→2⁺

 () =  (2).

6.The graph of  = () must have discontinuities at  = −1 and  = 4. It must show that

lim

→−1⁻ () =  (−1) and lim

→4⁺ () =  (4).