Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - Review Exercises - Page 124: 18

Answer

$$5$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{p \to 1} \frac{{{p^5} - 1}}{{p - 1}} \cr & {\text{Try to evaluate the limit}} \cr & \mathop {\lim }\limits_{p \to 1} \frac{{{p^5} - 1}}{{p - 1}} = \frac{{{{\left( 1 \right)}^5} - 1}}{{1 - 1}} = \frac{0}{0} \cr & {\text{Factor the numerator}} \cr & \mathop {\lim }\limits_{p \to 1} \frac{{{p^5} - 1}}{{p - 1}} = \mathop {\lim }\limits_{p \to 1} \frac{{\left( {p - 1} \right)\left( {{p^4} + {p^3} + {p^2} + p + 1} \right)}}{{p - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{p \to 1} \left( {{p^4} + {p^3} + {p^2} + p + 1} \right) \cr & {\text{Evaluate the limit}} \cr & \mathop {\lim }\limits_{p \to 1} \left( {{p^4} + {p^3} + {p^2} + p + 1} \right) = {1^4} + {1^3} + {1^2} + 1 + 1 \cr & \mathop {\lim }\limits_{p \to 1} \left( {{p^4} + {p^3} + {p^2} + p + 1} \right) = 5 \cr} $$
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