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4) Find the value of each limit: х* + х —6 а. lim х-я х — 2х-15 x→1 | Vx – 1 lim х—1 х — 1 b.

4) Find the value of each limit: х* + х —6 а. lim х-я х — 2х-15 x→1 | Vx – 1 lim х—1 х — 1 b.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**4) Find the value of each limit:** a. \(\lim_{{x \to 1}} \frac{{x^2 + x - 6}}{{x^2 - 2x - 15}}\) b. \(\lim_{{x \to 1}} \frac{{\sqrt{x} - 1}}{{x - 1}}\) --- **Explanation:** - **Problem part (a)**: The limit involves a rational function where both the numerator \(x^2 + x - 6\) and the denominator \(x^2 - 2x - 15\) are polynomials. To find the limit as \(x\) approaches 1, one might consider factoring both the numerator and the denominator and then simplifying the expression, if possible, to resolve the indeterminate form. - **Problem part (b)**: The limit features a quotient with a radical expression \(\sqrt{x} - 1\) in the numerator and \(x - 1\) in the denominator. This limit can often be tackled by rationalizing the numerator or applying L'Hôpital's rule if directly substituting the value results in an indeterminate form. --- **Educational context:** These types of limits are common in introductory calculus courses. They test one’s ability to manipulate algebraic expressions and understand the behavior of functions as a variable approaches a particular value. Mastery of these foundational skills is crucial for progressing in calculus and understanding more complex topics.
Transcribed Image Text:**4) Find the value of each limit:** a. \(\lim_{{x \to 1}} \frac{{x^2 + x - 6}}{{x^2 - 2x - 15}}\) b. \(\lim_{{x \to 1}} \frac{{\sqrt{x} - 1}}{{x - 1}}\) --- **Explanation:** - **Problem part (a)**: The limit involves a rational function where both the numerator \(x^2 + x - 6\) and the denominator \(x^2 - 2x - 15\) are polynomials. To find the limit as \(x\) approaches 1, one might consider factoring both the numerator and the denominator and then simplifying the expression, if possible, to resolve the indeterminate form. - **Problem part (b)**: The limit features a quotient with a radical expression \(\sqrt{x} - 1\) in the numerator and \(x - 1\) in the denominator. This limit can often be tackled by rationalizing the numerator or applying L'Hôpital's rule if directly substituting the value results in an indeterminate form. --- **Educational context:** These types of limits are common in introductory calculus courses. They test one’s ability to manipulate algebraic expressions and understand the behavior of functions as a variable approaches a particular value. Mastery of these foundational skills is crucial for progressing in calculus and understanding more complex topics.
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