Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.2 - Finding Limits Algebraically - 13.2 Exercises - Page 913: 17

Answer

7

Work Step by Step

Since there is no denominator, simply plug in 12. So $12^2 $ is 144 and 144 plus 25 is 169 and $\sqrt {169}=13$ and 13 minus $\sqrt {(3 \times 12)}$ is 13-6 or 7.
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