Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 33



Work Step by Step

We want to find $\lim\limits_{u \to 1}\dfrac{u^4-1}{u^3-1}$, but we can't use the quotient rule for limits because the limit of the denominator as $u$ approaches 1 is zero, and if we try to substitute $u=1$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{u^4-1}{u^3-1}=\dfrac{(u-1)(u^3+u^2+u+1)}{(u-1)(u^2+u+1)}=\dfrac{(u^3+u^2+u+1)}{u^2+u+1}$ for all $u \neq 1.$ Thus $\lim\limits_{u \to 1}\dfrac{u^4-1}{u^3-1}=\lim\limits_{u \to 1}\dfrac{u^3+u^2+u+1}{u^2+u+1}=\dfrac{1+1+1+1}{1+1+1}=\dfrac{4}{3}.$
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