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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 14: Q. 4 (page 907)

In Problems 1鈥� 6, find each limit
$\lim_{x 鈫� - 1} \frac{x^{2} - 4 x - 5}{x^{3} + 1}$

Short Answer

Expert verified

The value is $- 2$

Step by step solution

01

Step 1. Rewriting the limit

We can write the given limit in simplified form as,

$\lim_{x 鈫� - 1} \frac{x^{2} - 4 x - 5}{x^{3} + 1} = \lim_{x 鈫� - 1} \frac{(x + 1) (x - 5)}{(x + 1) (x^{2} - x + 1)}$

Cancelling common terms, the expression becomes,

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

02

Step 2. Value of the limit

On simplifying,

$\frac{\lim_{x 鈫� - 1}}{\lim_{x 鈫� - 1} (x^{2} - x + 1)} = \frac{\lim_{x 鈫� - 1} x - \lim_{x 鈫� - 1} 5}{\lim_{x 鈫� - 1} x^{2} - \lim_{x 鈫� - 1} x + \lim_{x 鈫� - 1} 1} = \frac{- 1 - \lim_{x 鈫� - 1} 5}{(- 1)^{2} - (- 1) + \lim_{x 鈫� - 1} 1} = \frac{- 1 - 5}{(- 1)^{2} - (- 1) + 1} = \frac{- 6}{3} = - 2$

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Most popular questions from this chapter

In Problems 7鈥� 42, find each limit algebraically.

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