Answered: x-2 51. lim x-1x-1

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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Find the one-sided limits

$### Problem 51 Evaluate the following limit as $ x $ approaches 1 from the left: \[ \lim_{{x \to 1^-}} \left( \frac{x - 2}{x - 1} \right) \] To solve the limit, consider the behavior of the function $\frac{x - 2}{x - 1}$ as $ x $ approaches 1 from the left. #### Solution: 1. **Substitute a value close to 1 from the left into the expression $\frac{x - 2}{x - 1}$:** - For $ x < 1 $, $(x - 1)$ is a small negative number. - Hence, $\frac{x - 2}{x - 1}$ evaluates the value. - As $ x $ gets closer to 1 from the left side, $(x - 2)$ approaches $-1$ and $(x - 1)$ approaches a very small negative number. 2. **Analyze the fraction:** - $\frac{x - 2}{x - 1} = \frac{-1}{\text{small negative number}}$ - Since dividing a negative number by another negative number results in a positive number, and the denominator is approaching zero. The value of the expression becomes very large positively. Thus, \[ \lim_{{x \to 1^-}} \left( \frac{x - 2}{x - 1} \right) = +\infty \] Therefore, the limit of $\frac{x - 2}{x - 1}$ as $ x $ approaches 1 from the left is $ +\infty $.$

Transcribed Image Text:### Problem 51 Evaluate the following limit as $ x $ approaches 1 from the left: \[ \lim_{{x \to 1^-}} \left( \frac{x - 2}{x - 1} \right) \] To solve the limit, consider the behavior of the function $\frac{x - 2}{x - 1}$ as $ x $ approaches 1 from the left. #### Solution: 1. **Substitute a value close to 1 from the left into the expression $\frac{x - 2}{x - 1}$:** - For $ x < 1 $, $(x - 1)$ is a small negative number. - Hence, $\frac{x - 2}{x - 1}$ evaluates the value. - As $ x $ gets closer to 1 from the left side, $(x - 2)$ approaches $-1$ and $(x - 1)$ approaches a very small negative number. 2. **Analyze the fraction:** - $\frac{x - 2}{x - 1} = \frac{-1}{\text{small negative number}}$ - Since dividing a negative number by another negative number results in a positive number, and the denominator is approaching zero. The value of the expression becomes very large positively. Thus, \[ \lim_{{x \to 1^-}} \left( \frac{x - 2}{x - 1} \right) = +\infty \] Therefore, the limit of $\frac{x - 2}{x - 1}$ as $ x $ approaches 1 from the left is $ +\infty $.

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