Section 2.6 Limits at Infinity; Horizontal Asymptotes

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Section 2.6 Limits at Infinity; Horizontal Asymptotes

30. Find the limit or show that it does not exist.

x→−∞lim

p4x²+ 3x + 2x . Solution:

SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 113

24. lim

→−∞

√1 + 4⁶

2 − ³ = lim

→−∞

√1 + 4⁶³ (2 − ³)³ =

→−∞lim −

(1 + 4⁶)⁶

→−∞lim (2³− 1) [since ³= −√

⁶for   0]

=

→−∞lim −

1⁶+ 4 2 lim

→−∞(1³) − lim_

→−∞1 =

−

→−∞lim (1⁶) + lim

→−∞4 2(0) − 1

= −

√0 + 4

−1 = −2

−1 = 2

25. lim

→∞

√ + 3² 4 − 1 = lim

→∞

√ + 3²

(4 − 1) =

lim→∞

( + 3²)²

lim→∞(4 − 1) [since  =√

²for   0]

=

lim→∞

1 + 3

lim→∞4 − lim__→∞(1) =

lim

→∞(1) + lim

→∞3

4 − 0 =

√0 + 3

4 =

√3 4

26. lim

→∞

 + 3² 4 − 1 = lim

→∞

( + 3²)

(4 − 1) = lim

→∞

1 + 3

4 − 1

= ∞ since 1 + 3 → ∞ and 4 − 1 → 4 as  → ∞.

27. lim

→∞

√9²+  − 3

= lim

→∞

√9²+  − 3√

9²+  + 3

√9²+  + 3 = lim

→∞

√9²+ 2

− (3)²

√9²+  + 3

= lim

→∞

9²+ 

− 9²

√9²+  + 3 = lim

→∞

√ 

9²+  + 3·1

1

= lim

→∞

 

9²²+ ²+ 3 = lim

→∞

 1

9 + 1 + 3 = 1

√9 + 3 = 1 3 + 3 = 1

6

28. lim

→−∞

√4²+ 3 + 2

= lim

→−∞

√4²+ 3 + 2 √

4²+ 3 − 2

√4²+ 3 − 2



= lim

→−∞

4²+ 3

− (2)²

√4²+ 3 − 2 = lim

→−∞

√ 3

4²+ 3 − 2

= lim

→−∞

√ 3

4²+ 3 − 2

 = lim

→−∞

3

−

4 + 3 − 2 [since  = −√

²for   0]

= 3

−√

4 + 0 − 2 = −3 4

29. lim

→∞

√²+  −√

²+ 

= lim

→∞

√²+  −√

²+  √

²+  +√

²+ 

√²+  +√

²+ 

= lim

→∞

(²+ ) − (²+ )

√²+  +√

²+  = lim

→∞

[( − )]

√²+  +√

²+ 

√

²

= lim

→∞

 − 

1 +  +

1 +  = √  −  1 + 0 +√

1 + 0 =  −  2

48. Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.

y = 2x²+ 1 3x²+ 2x − 1. Solution:

116 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 46. (a)

From the graph of

 () =√

3²+ 8 + 6 −√

3²+ 3 + 1, we estimate (to one decimal place) the value of lim

→∞ ()to be 14.

(b)

  ()

10,000 1443 39 100,000 1443 38 1,000,000 1443 38

From the table, we estimate (to four decimal places) the limit to be 14434.

(c) lim

→∞ () = lim

→∞

√3²+ 8 + 6 −√

3²+ 3 + 1√

3²+ 8 + 6 +√

3²+ 3 + 1

√3²+ 8 + 6 +√

3²+ 3 + 1

= lim

→∞

3²+ 8 + 6

−

3²+ 3 + 1

√3²+ 8 + 6 +√

3²+ 3 + 1 = lim

→∞

(5 + 5)(1)

√3²+ 8 + 6 +√

3²+ 3 + 1 (1)

= lim

→∞

5 + 5

3 + 8 + 6²+

3 + 3 + 1² = 5

√3 +√

3 = 5 2√

3 = 5√ 3

6 ≈ 1443376

47. lim

→±∞

5 + 4

 + 3 = lim

→±∞

(5 + 4)

( + 3) = lim

→±∞

5 + 4

1 + 3 = 0 + 4 1 + 0 = 4, so

 = 4is a horizontal asymptote.  = () = 5 + 4

 + 3, so lim

→−3⁺ () = −∞

since 5 + 4 → −7 and  + 3 → 0⁺as  → −3⁺. Thus,  = −3 is a vertical asymptote. The graph conﬁrms our work.

48. lim

→±∞

2²+ 1

3²+ 2 − 1= lim

→±∞

(2²+ 1)² (3²+ 2 − 1)²

= lim

→±∞

2 + 1²

3 + 2 − 1² = 2 3 so  = 2

3is a horizontal asymptote.  = () = 2²+ 1

3²+ 2 − 1 = 2²+ 1 (3 − 1)( + 1).

The denominator is zero when  = ¹₃and −1, but the numerator is nonzero, so  = ¹3and  = −1 are vertical asymptotes.

The graph conﬁrms our work.

49. lim

→±∞

2²+  − 1

²+  − 2 = lim

→±∞

2²+  − 1

²

²+  − 2

²

= lim

→±∞

2 + 1

− 1

² 1 + 1

− 2

²

=

→±∞lim

 2 + 1

− 1

²



→±∞lim

 1 + 1

− 2

²



=

→±∞lim 2 + lim

→±∞

1

− lim_

→±∞

1

²

→±∞lim 1 + lim

→±∞

1

− 2 lim_

→±∞

1

²

= 2 + 0 − 0

1 + 0 − 2(0) = 2, so  = 2 is a horizontal asymptote.

58. Find a formula for a function that has vertical asymptotes x = 1 and x = 3 and horizontal asymptote y = 1.

Solution:

SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 119 56.

(i)  = 0 (ii)   0 ( odd) (iii)   0 ( even) (iv)   0 ( odd) (v)   0 ( even) From these sketches we see that

(a) lim

→0⁺

^=







1 if  = 0 0 if   0

∞ if   0

(b) lim

→0⁻

^=











1 if  = 0 0 if   0

−∞ if   0,  odd

∞ if   0,  even

(c) lim

→∞^=







1 if  = 0

∞ if   0 0 if   0

(d) lim

→−∞^=











1 if  = 0

−∞ if   0,  odd

∞ if   0,  even 0 if   0

57. Let’s look for a rational function.

(1) lim

→±∞ () = 0 ⇒ degree of numerator  degree of denominator (2) lim

→0 () = −∞ ⇒ there is a factor of ²in the denominator (not just , since that would produce a sign change at  = 0), and the function is negative near  = 0.

(3) lim

→3⁻ () = ∞ and lim

→3⁺ () = −∞ ⇒ vertical asymptote at  = 3; there is a factor of ( − 3) in the denominator.

(4) (2) = 0 ⇒ 2 is an -intercept; there is at least one factor of ( − 2) 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

 () = 2 − 

²( − 3)as one possibility.

58. Since the function has vertical asymptotes  = 1 and  = 3, the denominator of the rational function we are looking for must have factors ( − 1) and ( − 3). Because the horizontal asymptote is  = 1, the degree of the numerator must equal the degree of the denominator, and the ratio of the leading coefﬁcients must be 1. One possibility is () = ²

( − 1)( − 3). 59. (a) We must ﬁrst ﬁnd the function . Since  has a vertical asymptote  = 4 and -intercept  = 1,  − 4 is a factor of the

denominator and  − 1 is a factor of the numerator. There is a removable discontinuity at  = −1, so  − (−1) =  + 1 is a factor of both the numerator and denominator. Thus,  now looks like this: () =( − 1)( + 1)

( − 4)( + 1) , where  is still to be determined. Then lim

→−1 () = lim

→−1

( − 1)( + 1)

( − 4)( + 1) = lim

→−1

( − 1)

 − 4 = (−1 − 1) (−1 − 4) = 2

5, so2

5 = 2, and

 = 5. Thus () = 5( − 1)( + 1)

( − 4)( + 1) is a ratio of quadratic functions satisfying all the given conditions and

 (0) =5(−1)(1) (−4)(1) = 5

4.

1