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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.4.59
Chapter 2, Problem 2.4.59

Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions.
f(x)=x^2−3x+2 / x^10−x^9

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1
Step 1: Identify the function's denominator, which is \( x^{10} - x^9 \). Set the denominator equal to zero to find the values of \( x \) that make the function undefined: \( x^{10} - x^9 = 0 \).
Step 2: Factor the denominator. Notice that \( x^{10} - x^9 \) can be factored as \( x^9(x - 1) = 0 \).
Step 3: Solve the factored equation \( x^9(x - 1) = 0 \) to find the critical points. This gives \( x^9 = 0 \) and \( x - 1 = 0 \).
Step 4: Solve \( x^9 = 0 \) to find \( x = 0 \). Solve \( x - 1 = 0 \) to find \( x = 1 \). These are the potential vertical asymptotes.
Step 5: Verify if these points are indeed vertical asymptotes by checking the behavior of the function as \( x \) approaches these values. Since the numerator \( x^2 - 3x + 2 \) does not have \( x = 0 \) or \( x = 1 \) as roots, both \( x = 0 \) and \( x = 1 \) are vertical asymptotes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Vertical Asymptotes

Vertical asymptotes occur in a function when the output approaches infinity as the input approaches a certain value. This typically happens at values of x that make the denominator of a rational function equal to zero, provided that the numerator does not also equal zero at those points. Identifying these points is crucial for understanding the behavior of the function near those values.
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Rational Functions

A rational function is a function that can be expressed as the ratio of two polynomials. In the given function f(x) = (x^2 - 3x + 2) / (x^10 - x^9), the numerator and denominator are both polynomials. Analyzing rational functions involves understanding how the degrees of the polynomials affect the function's behavior, particularly in terms of asymptotes and intercepts.
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Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler components, or factors, that can be multiplied together to yield the original polynomial. This is essential for identifying the roots of the numerator and denominator in rational functions, which helps in determining vertical asymptotes and simplifying the function for analysis.
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Related Practice
Textbook Question

Consider the position function s(t) =−16t^2+100t representing the position of an object moving vertically along a line. Sketch a graph of s with the secant line passing through (0.5, s(0.5)) and (2, s(2)). Determine the slope of the secant line and explain its relationship to the moving object.

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Textbook Question

Sketch a graph of f and use it to make a conjecture about the values of f(a), lim x→a^−f(x),lim x→a^+f(x), and lim x→a f(x) or state that they do not exist.

f(x) = x^2+x−2 / x−1; a=1

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Textbook Question

Use the precise definition of infinite limits to prove the following limits.


limx→41(x−4)2=∞{\(\displaystyle\]\lim\)_{x\(\to\)4}}\(\frac{1}{\left(x-4\right)^2}\)=\(\infty\)

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Textbook Question

Sketch a possible graph of a function g, together with vertical asymptotes, satisfying all the following conditions.


g(2) =1,g(5) =−1,lim x→4 g(x) =−∞,lim x→7^− g(x) =∞,lim x→7^+ g(x) =−∞

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Textbook Question

A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence {2,4,6,8,…} is specified by the function f(n) = 2n, where n=1,2,3,….The limit of such a sequence is lim n→∞ f(n), provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences or state that the limit does not exist. 


{0,1/2,2/3,3/4,…}, which is defined by f(n) = (n−1) / n, for n=1,2,3,…

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Textbook Question

Use an appropriate limit definition to prove the following limits.


lim x→ 5x^2 − 25 / x − 5=10

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