Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 73: 41


$$\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}=\frac{4}{3}$$

Work Step by Step

Given $$ \lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}$$ let $$ f(x) = \frac{x^{4}-1}{x^{3}-1}$$ Since, we have $$ f( 1)= \frac{1-1}{1-1}=\frac{0}{0}$$ So, transform algebraically and cancel, we get \begin{aligned} L&= \lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}\\ &=\lim _{x \rightarrow 1} \frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{(x-1)\left(x^{2}+x+1\right)} \\ &=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)\left(x^{2}+1\right)}{(x-1)\left(x^{2}+x+1\right)}\\ &=\lim _{x \rightarrow 1} \frac{(x+1)\left(x^{2}+1\right)}{\left(x^{2}+x+1\right)}\\ &=\frac{4}{3}\\ \end{aligned}
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