## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 56: 28

#### Answer

$-\dfrac{1}{2}$

#### Work Step by Step

We want to find $\lim\limits_{t \to -1}\dfrac{t^2+3t+2}{t^2-t-2}$, 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 $\dfrac{t^2+3t+2}{t^2-t-2}=\dfrac{(t+2)(t+1)}{(t-2)(t+1)}=\dfrac{t+2}{t-1}$ for all $t \neq -1.$ Thus $\lim\limits_{t \to -1}\dfrac{t^2+3t+2}{t^2-t-2}=\lim\limits_{t \to -1}\dfrac{t+2}{t-1}=\dfrac{-1+2}{-1-1}=\dfrac{1}{-2}=-\dfrac{1}{2}.$

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