Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problem 2.2.37

Chapter 2, Problem 2.2.37

Limits of quotients

Find the limits in Exercises 23–42.

limx→1 (x −1) / (√(x + 3) − 2)

Verified step by step guidance

Identify the form of the limit: As x approaches 1, both the numerator (x - 1) and the denominator (√(x + 3) - 2) approach 0, indicating an indeterminate form 0/0.

To resolve the indeterminate form, consider rationalizing the denominator. Multiply the numerator and the denominator by the conjugate of the denominator: (√(x + 3) + 2).

Rewrite the expression: ((x - 1) * (√(x + 3) + 2)) / ((√(x + 3) - 2) * (√(x + 3) + 2)).

Simplify the denominator using the difference of squares formula: (√(x + 3))^2 - 2^2 = x + 3 - 4 = x - 1.

Cancel the common factor (x - 1) in the numerator and denominator, then evaluate the limit of the simplified expression as x approaches 1.

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Key Concepts

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

Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. They help in understanding the behavior of functions near specific points, especially when direct substitution leads to indeterminate forms. In this case, evaluating the limit as x approaches 1 requires careful manipulation to avoid division by zero.

Quotient of Functions

The quotient of functions involves dividing one function by another. When finding limits of quotients, it is essential to analyze both the numerator and denominator to determine their behavior as the variable approaches a specific value. In this exercise, the expression (x - 1) in the numerator and the square root function in the denominator must be examined to simplify the limit.

Rationalization

Rationalization is a technique used to eliminate radicals from the denominator of a fraction. This is particularly useful when evaluating limits that result in indeterminate forms. In the given limit, multiplying the numerator and denominator by the conjugate of the denominator, √(x + 3) + 2, can simplify the expression and make it easier to evaluate the limit as x approaches 1.

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Related Practice

Textbook Question

Using the Sandwich Theorem

a. Suppose that the inequalities 1/2 − x² / 24 < (1 − cos x)/ x² < 1/2 hold for values of x close to zero, except for x = 0 itself. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about limx→0 (1 −cos x)/ x²?

Give reasons for your answer.

[Technology Exercise] b. Graph the equations y=(1/2) − (x²/24), y = (1 - cos x) / x², and y = 1/2 together for −2 ≤ x ≤2. Comment on the behavior of the graphs as x→0.

Textbook Question

Use formal definitions to prove the limit statements in Exercises 93–96.

lim x → −5 (1 / (x + 5)²) = ∞

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.)

g(x) = 1/(2 + (1/x))

Textbook Question

Limits and Infinity

Find the limits in Exercises 37–46.

sin x

lim ------------- ( If you have a grapher, try graphing

x→∞ |x| the function for ―5 ≤ x ≤ 5 ) .

Textbook Question

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limθ→0 (1 − cos θ) / sin 2θ

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