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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 2.2.51
Chapter 2, Problem 2.2.51

Using Limit Rules


Suppose lim x→0 f(x) = 1 and lim x→0 g(x) = −5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.


limx→0 (2f(x) − g(x)) / (f(x) + 7)² = limx→0 (2f(x) − g(x)) / limx→0 (f(x) + 7)² (a)


(We assume the denominator is nonzero.)


(lim x→0 2f(x) − lim x→0 g(x)) / (lim x→0 (f(x) + 7))² (b)


= (2 lim x→0 f(x) − lim x→0 g(x)) / (lim x→0 f(x) + lim x→0 7)² (c)


= ((2)(1) − (−5)) / (1 + 7)² = 7/64

Verified step by step guidance
1
Step 1: Identify the limit expression given: lim_{x→0} (2f(x) − g(x)) / (f(x) + 7)². We need to apply limit rules to simplify this expression.
Step 2: Apply the Quotient Rule for limits, which states that lim_{x→a} [u(x)/v(x)] = [lim_{x→a} u(x)] / [lim_{x→a} v(x)], provided lim_{x→a} v(x) ≠ 0. This allows us to separate the limit of the numerator and the denominator.
Step 3: For the numerator, apply the Sum/Difference Rule for limits: lim_{x→a} [u(x) ± v(x)] = lim_{x→a} u(x) ± lim_{x→a} v(x). This lets us separate the terms 2f(x) and -g(x) into individual limits.
Step 4: For the term 2f(x), apply the Constant Multiple Rule: lim_{x→a} [c * u(x)] = c * lim_{x→a} u(x). This allows us to take the constant 2 out of the limit.
Step 5: For the denominator, apply the Sum Rule and the Constant Rule: lim_{x→a} [u(x) + c] = lim_{x→a} u(x) + c, where c is a constant. This simplifies the denominator to (lim_{x→0} f(x) + 7)².

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Key Concepts

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

Limit of a Function

The limit of a function describes the value that a function approaches as the input approaches a certain point. In this case, we have limits for f(x) and g(x) as x approaches 0, which are essential for evaluating the overall limit of the expression. Understanding limits is fundamental in calculus as it lays the groundwork for concepts like continuity and derivatives.
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Limits of Rational Functions: Denominator = 0

Limit Laws

Limit laws are a set of rules that allow us to manipulate limits algebraically. These include properties such as the sum, difference, product, and quotient of limits. In the given problem, these laws are applied to break down the limit of a complex expression into simpler parts, making it easier to evaluate the limit as x approaches 0.
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One-Sided Limits

Continuity and Non-zero Denominator

For a limit to be evaluated using the quotient rule, the denominator must be non-zero at the point of interest. Continuity ensures that the function behaves predictably around that point. In this problem, it is assumed that the denominator (f(x) + 7)² does not approach zero as x approaches 0, allowing the application of limit laws without encountering undefined behavior.
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Intro to Continuity
Related Practice
Textbook Question

Finding Limits


In Exercises 3–8, find the limit of each function (a) as x → ∞ and (b) as x → −∞. (You may wish to visualize your answer with a graphing calculator or computer.)


h(x) = (−5 + (7/x))/(3 – (1/x²))

Textbook Question

Using the Formal Definitions


Use the formal definitions of limits as x → ±∞ to establish the limits in Exercises 91 and 92.


If f has the constant value f(x) = k, then lim x → ∞ f(x) = k.

Textbook Question

Define h(2) in a way that extends h(t) = (t² + 3t − 10)/(t − 2) to be continuous at t = 2.

Textbook Question

For what values of a and b is

g(x) = { ax + 2b, x ≤ 0

x² + 3a – b, 0 < x ≤ 2

3x – 5, x > 2

continuous at every x?

1
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Textbook Question

Limits as x → ∞ or x → −∞


The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x. Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36. Write ∞ or −∞ where appropriate.


lim x → ⁻∞ ((1 − x³) / (x² + 7x))⁵

Textbook Question

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


limx→9 √(x − 5) = 2