Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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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.
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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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