Calculus 8th Edition answers to Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises - Page 91 2 including work step by step written by community members like you. Textbook Authors: Stewart, James, ISBN-10:, ISBN-13: 978-1-28574-062-1, Publisher: Cengage.
Master AP Calculus AB & BC
Worksheet 1.3—Limits at Infinity Show all work. No calculator Short Answer: On problems 1 – 6, find (a) lim ( ) x f x →∞ (b) lim ( ) x f x →−∞ (c) the equations of any horizontal or slant asymptotes. 3 2 2 3 4 1 13 x x x f x x x − − − = + − 2. 2 3 4 3 5 2 1 x x f x x x − + = + − 3. 3 1 4 x f x x + = − 4. 750 Chapter 11 Limits and an Introduction to Calculus The Limit Concept The notion of a limit is a fundamental concept of calculus. In this chapter, you will learn how to evaluate limits and how they are used in the two basic problems of calculus: the tangent line problem and the area problem. Example 1 Finding a Rectangle of Maximum Area. Here, our limit as x approaches infinity is still two, but our limit as x approaches negative infinity, right over here, would be negative two. And of course, there's many situations where, as you approach infinity or negative infinity, you aren't actually approaching some finite value. Section 4-11: Linear Approximations. In this section we’re going to take a look at an application not of derivatives but of the tangent line to a function.
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 3. Limits and Continuity
LIMITS INVOLVING INFINITY
You can learn a lot about a function from its asymptotes, so it’s important that you can determine what kind of asymptotes shape a graph just by looking at a function. Remember that asymptotes are lines that a graph approaches but never reaches, as the graph stretches out forever. The two kinds of asymptotes with which you should concern yourself are vertical asymptotes and horizontal asymptotes; in the graph below of g(x), x = —2 is a vertical asymptote and y = 4 is a horizontal asymptote.
Some students are confused by this diagram, since g(x) actually intersects the horizontal asymptote. “I thought a graph can’t hit an asymptote,” they mutter, eyes filling with tears. A graph can intersect its asymptote, as long as it doesn’t make a habit of it. Even though g intersects y = 4 at (2,4), g only gets closer and closer toy = 4 after that (for x > 2), and g won’t intersect the line out there near infinity—it’s the infinite behavior of the function that concerns us. It’s the same with the criminal justice system—if you cross paths with the law a couple of times when you’re very young, it’s not that big a deal, but as you get much older, the police tend to frown upon your crossing them again.
Vertical asymptotes are discontinuities that force a function to increase or decrease without bound to avoid an x value. For example, consider the graph of
In this case, the function has an infinite discontinuity at x = 1. As you approach x = 1 from the left, the function decreases without bound, and from the right, you increase without bound. In general, if f(x) has a vertical asymptote at x = c, then or —∞. This is commonly called an infinite limit. This terminology is slightly confusing, because when/has an infinite limit at c, f has no limit at c!
Example 7: Determine which discontinuities of are caused by vertical asymptotes.
Solution: To begin, factor p(x) to get Because the denominator of a fraction cannot equal zero, p is discontinuous at x = —2 and —3. However, using the factoring method, exists and equals Therefore, the discontinuity at x = —2 is removable and not a vertical asymptote; however, x = —3 is a vertical asymptote. If you substitute x = —3 into the p, you will get —1/0. Remember, a constant divided by zero is the fingerprint of a vertical asymptote.
Horizontal asymptotes (or limits at infinity) are limiting heights that a graph approaches as x gets infinitely large or small. Consider the graph below of
As x gets infinitely large (the extreme right side of the graph), the function is approaching a height of 2; in fact, the same is true as x gets infinitely negative (the left side of the graph). In this case, we write The AP Calculus test often features problems of the type and there is a handy trick to finding these limits at infinity of rational functions. We’ll begin with a generic example to learn the technique.
TIP. A rational function will always approach the same limit as x→∞ and x→—∞.
Example 8: Evaluate
Solution: Let A be the degree of f(x) and B be the degree of g(x).
• If A > B, then the limit is ∞.
• If B > A, then the limit is 0.
• If A = B, then the limit is the ratio of the leading coefficients of f(x) and g(x).
This technique only works for rational functions when you are finding the limit as x approaches infinity, and although it may sound tricky at first, the method is quite easy in practice.
NOTE. Remember that the degree of a polynomial is its highest exponent, and the coefficient of the term with the highest exponent is called the leading coefficient.
Example 9: Evaluate the following limits.
Because the degree of the numerator is greater than the degree of the denominator (4 > 2), the limit is ∞. In other words, the function does not approach a limiting height and will reach higher and higher as x increases.
The degree of the numerator is 3, since and the degree of the denominator is 5. Since the denominator’s degree is higher, the limit is 0.
The degrees of the numerator and denominator are both 4, so take the ratio of those terms’ coefficients to get a limit of —7/5.
Example 10: Give the equations of the vertical and horizontal asymptotes of
NOTE. Remember: If limit equals ∞, there technically is no limit, since ∞ is not a real number.
Solution: The horizontal asymptotes are easy to find using the technique of the previous two examples. The degrees of the numerator and denominator are equal, so Therefore, k has horizontal asymptote y = 2. In order to find any vertical asymptotes, begin by factoring k:
From this, you can see that k has discontinuities at x = 4 and 8. To determine which represents an asymptote, substitute them both into k. k(4) = 0/0 and k(8) = 76/0. Thus, x = 8 is an infinite discontinuity, and x = 4 is a point discontinuity, as verified by the graph.
EXERCISE 6
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
1. Explain how horizontal and vertical asymptotes are related to infinity.
2. If m is an even function with vertical asymptote x = 2 and draw a possible graph of m(x).
3. Evaluate:
4. Given a and b are positive integers, f has horizontal asymptote y = 2, and f has vertical asymptote x = 3:
Smashingfree flash games. (a) Find the correct values of a and b.
(b) Find the point of removable discontinuity on f.
5. Draw a function g(x) that satisfies all of the following properties:
• Domain of g is (—∞,—2) ∪ (—2,—∞)
• g has a nonremovable discontinuity at x = —2
• Range of g is [—3,∞)
• g has one root: x = —4
ANSWERS AND EXPLANATIONS
1. When a function approaches the vertical asymptote x = c, the function values either increase or decrease without bound (infinitely); for example: A horizontal asymptote occurs when a function approaches a fixed height forever as x approaches infinity (for example: ).
2. Because m is even, it must be y-symmetric, and, therefore, have a vertical asymptote at x = —2 as well. Furthermore, must also equal 0. Any solution must fit those characteristics; here is one:
3. (a) Because the degree of the denominator is greater than the degree of the numerator, the limit at infinity is zero.
(b) The numerator and denominator are both of degree 3 (since ), so take the corresponding coefficients to find the limit of . Note that the radical remains around the 7.
4. (a) If / has vertical asymptote x = 3, we can find a. Remember the vertical asymptote fingerprint: a zero in the denominator but not in the numerator. The denominator equals zero when 32 — a2= 0.
1.8 Limits With Infinityap Calculus Answers
32 - a2 = 0
a2 = 9
a = 3 (since a has to be positive according to the problem)
Substitute a into the fraction to give you
If f has horizontal asymptote y = 2, then The numerator and denominator have the same degree, so the limit is b/2 = 2. Thus, b = 4.
(b) Substituting both b and a gives you
Factor completely to get
The (x + 3) factor can be eliminated, meaning that Since x = —3 is a discontinuity at which a limit exists, f is removably discontinuous at the point (—3,5/6).
1.8 Limits With Infinityap Calculus Pdf
5. This problem is pretty involved, and all solutions will look similar to the graph below.
In order to get the range of [—3,∞), it’s important that The graph must also reach down to and include the height of —3, although it need not happen at (—6, —3) as on this graph.
Inverse functions
Recall that a function (f ) is one-to-one (often written as (1-1)) if it assigns distinct values of (y ) to distinct values of (x ). In other words, if (x_1 ne x_2 ) then (f(x_1 ) ne f(x_2 ) ). Equivalently, (f ) is one-to-one if (f(x_1 ) = f(x_2 ) ) implies (x_1 = x_2 ). There is a simple horizontal rule for determining whether a function (y=f(x) ) is one-to-one: (f ) is one-to-one if and only if every horizontal line intersects the graph of (y=f(x) ) in the (xy)-coordinate plane at most once (see Figure 5.3.3).
Figure 5.3.3 Horizontal rule for one-to-one functions
If a function (f ) is one-to-one on its domain, then (f ) has an inverse function, denoted by (f^{-1} ), such that (y=f(x) ) if and only if (f^{-1}(y) = x ). The domain of (f^{-1} ) is the range of (f ).
The basic idea is that (f^{-1} ) 'undoes' what (f ) does, and vice versa. In other words,
[nonumber begin{alignat*}{3}
f^{-1}(f(x)) ~&=~ x quad&&text{for all (x ) in the domain of (f ), and} nonumber
f(f^{-1}(y)) ~&=~ y quad&&text{for all (y ) in the range of (f ).}
end{alignat*}]
Theorem (PageIndex{1})
If (f) is continuous and one to one, then (f^{-1} is continuous on its domain. 5 things to do when your products appear on social media.