Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 20



Work Step by Step

By substituting, we get the intermediate form $\frac{0}{0}$. So we multiply the function by $ \frac{1+\sqrt{S^2+1}}{1+\sqrt{S^2+1}}$; then we have $$ \lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2}=\lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2} \frac{1+\sqrt{S^2+1}}{1+\sqrt{S^2+1}}\\ =\lim _{s \rightarrow 0} \frac{1-(S^2+1) }{s^2(1+\sqrt{S^2+1})} =\lim _{s \rightarrow 0} \frac{-1 }{ 1+\sqrt{S^2+1}}= \frac{-1}{1+1}=\frac{-1}{2}. $$
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