Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 27

Answer

$$1$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{x} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + x} - \sqrt {1 - x} } \right) = \sqrt {1 + 0} - \sqrt {1 - 0} = 1 - 1 = 0 \cr & \mathop {\lim }\limits_{x \to 0} x = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}}{\text{.}} \cr & {\text{for }}\sqrt {1 + x} - \sqrt {1 - x} \to {D_x}\left( {\sqrt {1 + x} - \sqrt {1 - x} } \right) = \frac{1}{{2\sqrt {1 + x} }} - \frac{{ - 1}}{{2\sqrt {1 - x} }} \cr & {\text{for }}x \to {D_x}\left( x \right) = 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }}} \right) \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{2\sqrt {1 + 0} }} + \frac{1}{{2\sqrt {1 - 0} }} \cr & = \frac{1}{2} + \frac{1}{2} \cr & = 1 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{x} = 1 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.