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: 27

Answer

$\dfrac{3}{2}$

Work Step by Step

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