Limits

Calculus ยท 653 exercises

Q. 56

Calculate each limit in Exercises 35–80. 

limxx72-x83x2

2 step solution

Q. 57

Calculate each limit in Exercises 35–80. 

limx4x-3x5x

2 step solution

Q. 58

Calculate each limit in Exercises 35–80. 

limx2x-4-x3x

2 step solution

Q. 59

Calculate each limit in Exercises 35–80.  

limx-3x-5x4x

2 step solution

Q. 60

Calculate each limit in Exercises 35–80.  

limx4(3x)2+3x

2 step solution

Q. 61

Calculate each limit in Exercises 35–80.  

limx2e1.5x3e2x+e1.5x

2 step solution

Q. 62

Calculate each limit in Exercises 35–80.   

limx1-5e2x3ex+4e2x

2 step solution

Q. 63

Calculate each limit in Exercises 35–80. 

limx3+ln(x2-9)

2 step solution

Q. 64

Calculate each limit in Exercises 35–80. 

limx0+ln1x

2 step solution

Q. 65

Calculate each limit in Exercises 35–80. 

limxlnx2-ln(2x+1)

2 step solution

Q. 66

Calculate each limit in Exercises 35–80. 

limx(ln3x-ln2x)

2 step solution

Q. 67

Calculate each limit in Exercises 35–80. 

limx01-cos2x7x

2 step solution

Q. 68

Calculate each limit in Exercises 35–80. 

limx0sin3x5x

2 step solution

Q. 69

Calculate each limit in Exercises 35–80. 

limx0x1-cosx

2 step solution

Q. 70

Calculate each limit in Exercises 35–80. 

limx03sinx+xx

2 step solution

Q. 71

Calculate each limit in Exercises 35–80. 

limx0sin23xx3-x

2 step solution

Q. 72

Calculate each limit in Exercises 35–80. 

limx0sin(3x2)x3-x

2 step solution

Q 73.

Calculate the limit.

limx0+ x2csc 3x1-cos 2x

2 step solution

Q 74.

Find the limit.

limx0x2cot xsin x

2 step solution

Q 75.

Find the limit.

limx0sec x tan xx

2 step solution

Q 76.

Find the limit.

limx03x2 cot2x 

2 step solution

Q 77.

Find the limit.

limx01+x2x

2 step solution

Q 78.

Find the limit.

limx01+2x3x

2 step solution

Q 79.

Find the limit.

limx1+1x3x

2 step solution

Q 80.

Find the limit.

limx1-5xx

2 step solution

Q.48

Calculate each limit in Exercises 35–80. 

limx1 x-1x2-2x+1

2 step solution

Q 81.

In 1960, H. von Foerster suggested that the human population could be measured by the function Pt=179×1092027-t0.99

The time t is measured in years, where t = 1 corresponds to the year 1 a.d., 

t = 1973 corresponds to the year 1973 a.d., and so on. (We saw this “doomsday model” for population in Problem 77 of Section 1.1, on page 89.) Use limit

techniques to calculate limt2027-Pt. What does this limit mean in real-world terms?


2 step solution

Q 82.

Suppose instead we consider the population model

Qt=44×10101+2027-t43

with t measured in years as in the previous problem.

(a) Use limit techniques to calculate limtQt. What does this limit mean in real–world terms? What happens in this model in the year 2027?

(b) Use calculator graphs to compare the population models in this exercise with those in the previous exercise. Describe the long–term population growth

scenarios that are suggested by these models.

3 step solution

Q. 83

Consider a mass hanging from the ceiling at the end of a spring. If you pull down on the mass and let go, it will oscillate up and down according to the equation

s(t)=Asinkmt+Bcoskmt

where s(t) is the distance of the mass from its equilibrium position, m is the mass of the bob on the end of the spring, and k is a "spring coefficient" that measures how tight or stiff the spring is. The constants A and B depend on initial conditions - specifically, how far you pull down the mass s0 and the velocity at which you release the mass v0. This equation does not take into effect any friction due to air resistance.

  1. Determine whether or not the limit of s(t) as t exists. What does this say about the long-term behavior of the mass on the end of the spring?
  2. Explain how this limit relates to the fact that the equation for s(t) does not take friction due to air resistance into account.
  3. Suppose the bob at the end of the spring has a mass of 2 grams and that the coefficient for the spring is k=9. Suppose also that the spring is released in such a way that A=2 and B=2. Use a graphing utility to graph the function s(t) that describes the distance of the mass from its equilibrium position. Use your graph to support your answer to part (a).

4 step solution

Q. 84

In the previous exercise we gave an equation describing spring motion without air resistance. If we take into account friction due to air resistance, the mass will oscillate up and down according to the equation

s(t)=e(-f/2m)tAsin4km-f22mt+Bcos4km-f22mt

where m, k, A, and B are the constants described in Problem 83 and f is a positive "friction coefficient" that measures the amount of friction due to air resistance.

  1. Find the limit of s(t) as t. What does this say about the long-term behavior of the mass on the end of the spring?
  2. Explain how this limit relates to the fact that the new equation for s(t) does take friction due to air resistance into account.
  3. Suppose the bob at the end of the spring has a mass of 2 grams, the coefficient for the spring is k=9, and the friction coefficient is f=6. Suppose also that the spring is released in such a way that A=4 and B=2. Use a graphing utility to graph the function s(t) that describes the distance of the mass from its equilibrium position. Use your graph to support your answer to part (a).

4 step solution

Q. 1

A limit representing an instantaneous rate of change: After t seconds, a bowling ball dropped from 350 feet has height h(t)=350-16t2,measured in feet.

Calculate the average rate of change of the height of the bowling ball from t=3 to t=3+h seconds in the cases where h is equal to 0.5, 0.25, 0.1, and0.01.

3 step solution

Q. 2

A limit representing an instantaneous rate of change: After t seconds, a bowling ball dropped from 350 feet has height h(t)=350-16t2,measured in feet.

Write down a formula for the average rate of change of the height of the bowling ball from time t=3 to time t=3+h, assuming that h>0. The only letter in your formula should be h.


2 step solution

Q. 3

A limit representing an instantaneous rate of change: After t seconds, a bowling ball dropped from 350 feet has height h(t)=350-16t2,measured in feet.

Take the limit as h0+ of the formula you found for average rate of change in the previous problem. What does this limit represent in real-world terms?

2 step solution

Q. 4

In this section we learned that e can be thought of as the following limit:

limh01+h1h=e.

In the following exercise, you will investigate the convergence of this limit and also get a preview of the Taylor series, which we will see in Chapter 8.

Use the substitution n=1h to show that the preceding limit statement is equivalent to the limit statement

limn1+1nn=e.

2 step solution

Q. 5

The Binomial Theorem says that an expression of the form a+bncan be expanded to n0anb0+n1an-1b1+n2an-1b2++nnx0yn,where for any 0kn, the symbol knis equal to n!k!n-k!. Here n! is n factorial, the product of the integers from 1 to n. By convention we set 0!=1. Apply this expansion to the expression 1+1nn.

2 step solution

Q. 6

Show that as n we would expect the preceding expansion to approach

1+11!+12!+13!+14!+15!+.

3 step solution

Q. 7

Use a calculator to find the sum of the first six terms of the sum from the previous problem, and compare this sum with your calculator’s best approximation of the number e.

2 step solution

Q. 85

Use limits to prove that the limits of a polynomial f(x)=anxn+an-1xn-1+a1x+a0 are the same as the limits of its leading term anxn as x and as x-. (Hint: Show that limxf(x) is equal to limxanxn by factoring out anxn from f(x).)

2 step solution

Q. 86

Use limit techniques to prove that a rational function f(x)=p(x)q(x) will have,

  1. a horizontal asymptote at y = 0, if the degree of p(x) is less than the degree of q(x);
  2. a horizontal asymptote at y=anbm, where an and bnare the leading terms of p(x) and q(x), respectively, if p(x) and q(x) have the same degree;
  3. no horizontal asymptote, if the degree of p(x) is greater than the degree of q(x).

4 step solution

Q. 87

Prove the second part of Theorem 1.29: If limxcf(x)g(x) is of the form 10-, then limxcf(x)g(x)=-.

2 step solution

Q. 88

Prove the second part of Theorem 1.30: If limxf(x)g(x) is of the form 1-, then limxf(x)g(x)=0.

2 step solution

Q. 89

Prove that the sum rule for limits also applies for limits as x : If limxf(x)=L and limxg(x)=M, then limx(f(x)+g(x))=L+M.

2 step solution

Q. 91

Prove the second part of Theorem 1.31(a): If k>0, then limxx-k=0.

2 step solution

Q. 92

Prove the k=1 case of the first part of Theorem 1.31(b): that limxex=. (Hint: Given M>0, choose N=lnM. Then if x>N=lnM, we must have x=lnM+c for some positive number c. Use this to show that ex>M.)

2 step solution

Q. 93

Prove the k=1 case of the second part of Theorem 1.31(b): that limxe-x=0.

2 step solution

Q. 94

Prove that limθ01-cosθθ=0 by using the double-angle identity cos2θ=1-2sin2θ and the other special trigonometric limit limθ0sinθθ=1.

2 step solution

Q. 1

Give precise mathematical definitions or descriptions of each of the concepts that follow. Then illustrate the definition with a graph or algebraic example, if possible. 

the intuitive meaning of the limit statements limxcf(x)=L,limxc-f(x)=Land limxc+f(x)=L

4 step solution

Q. 1

Fill in the blanks to complete each of the following theorem statements: 

If limxcf(x)=Landlimxcf(x)=M, then                  

2 step solution

Q. 2

Give precise mathematical definitions or descriptions of each of the concepts that follow. Then illustrate the definition with a graph or algebraic example, if possible. 

the intuitive meaning of the limit statements limxcf(x)= and limxf(x)=L

3 step solution

Q. 2

Fill in the blanks to complete each of the following theorem statements: 

limxcf(x)=L if and only if limxc-f(x)=             andlimxc+f(x)=            .

2 step solution

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