Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2: 25

Answer

$\lim\limits_{x \to 4^{-}}\frac{3-x}{x²-2x-8}$ = +$\infty$

Work Step by Step

x² - 2x - 8 = (x - 4)(x + 2), so: $\lim\limits_{x \to 4^{-}}\frac{3-x}{x²-2x-8}$ = $\lim\limits_{x \to 4^{-}}\frac{3-x}{(x - 4)(x + 2)}$ = $\frac{3-4^{4}}{(4^{-}-4)(4^{-}+2)}$ = $\frac{-1}{0^{-}\times6}$ = $\frac{-1}{0^{-}}$ = +$\infty$
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