Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 53

Answer

$$3$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] \cr & {\text{Evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] = 3{\left( 0 \right)^{0/2}} = {0^0} \cr & {\text{This limit has the form }}{0^0}{\text{ }}\left( {{\text{See example 6, page 562}}} \right) \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] \cr & {\text{Take the natural log of each side}} \cr & \ln y = \ln \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] \cr & {\text{Continuity}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\ln \left[ {3{{\left( x \right)}^{x/2}}} \right]} \right) \cr & {\text{Product property of logarithms}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3 + \ln {{\left( x \right)}^{x/2}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3 + \frac{x}{2}\ln \left( x \right)} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3} \right] + \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{x}{2}\ln \left( x \right)} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3} \right] + \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\ln x}}{{2/x}}} \right] \cr & {\text{Evaluating}} \cr & \ln y = \ln 3 + \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3} \right] + \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\frac{d}{{dx}}\left[ {\ln x} \right]}}{{\frac{d}{{dx}}\left[ {2/x} \right]}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3} \right] + \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1/x}}{{ - 2/{x^2}}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\ln 3} \right] - \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{2} \cr & {\text{Evaluating the limit}} \cr & \ln y = \ln 3 - \frac{0}{2} \cr & \ln y = \ln 3 \cr & y = 3 \cr & {\text{Therefore}} \cr & y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] \cr & \mathop {\lim }\limits_{x \to {0^ + }} \left[ {3{{\left( x \right)}^{x/2}}} \right] = 3 \cr} $$
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