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Chapter 1: Q. 7 (page 119)

Given the following function $f$ , define $f (1)$ so that $f$ is continuous at x = 1, if possible:
$f (x) = \frac{x^{2} - 2 x + 1}{x^{2} - 6 x + 5}$

Short Answer

Expert verified

After taking limit

$\lim_{x 鈫� 1} \frac{x^{2} - 2 x + 1}{x^{2} - 6 x + 5} = 0$

Step by step solution

01

Step 1. Given information

Function $f (x) = \frac{x^{2} - 2 x + 1}{x^{2} - 6 x + 5}$

02

Step 2. Taking limit in the function

$\lim_{x 鈫� 1} f (x) = \lim_{x 鈫� 1} \frac{x^{2} - 2 x + 1}{x^{2} - 6 x + 5} = \lim_{x 鈫� 1} \frac{{(x - 1)}^{2}}{x^{2} - 5 x - x + 5} = \lim_{x 鈫� 1} \frac{{(x - 1)}^{2}}{x (x - 5) - 1 (x - 5)} = \lim_{x 鈫� 1} \frac{{(x - 1)}^{2}}{(x - 5) (x - 1)} = \lim_{x 鈫� 1} \frac{(x - 1)}{(x - 5)} = \frac{\lim_{x 鈫� 1} (x - 1)}{\lim_{x 鈫� 1} (x - 5)} = \frac{1 - 1}{1 - 5} = \frac{0}{- 4} = 0$

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Most popular questions from this chapter

Sketch a labeled graph of a function that satisfies the hypothesis of the Intermediate Value Theorem, and illustrate on your graph that the conclusion of the Intermediate Value Theorem follows.

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