Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1154: 59

Answer

a) The limit $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L $ for $ L={{L}_{0}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}$ is 0. b) If a starship is moving at a velocity approaching the speed of light, then its length appears to approach zero, because $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=0$. c) The reason for using the left hand limit in part (a), $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L $ is that the speed of the starship cannot exceed the speed of light.

Work Step by Step

(a) Consider the provided limit, $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L $. Substitute $ L={{L}_{0}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}$: $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\cdot g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$: $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\ & =\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \end{align}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,\sqrt{f\left( x \right)}=\sqrt{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}$ in $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}$: $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\ & =\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\ & =\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)} \end{align}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in $\sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)}$: $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)} \\ & =\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,1-\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\frac{{{v}^{2}}}{{{c}^{2}}}} \end{align}$ Use quotient property of limits in $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\frac{{{v}^{2}}}{{{c}^{2}}}$: $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,1-\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,\frac{{{v}^{2}}}{{{c}^{2}}}} \\ & =\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,1-\frac{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{v}^{2}}}{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{c}^{2}}}} \end{align}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,c=c,$ $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{L}_{0}}\cdot \sqrt{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,1-\frac{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{v}^{2}}}{\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,{{c}^{2}}}} \\ & ={{L}_{0}}\cdot \sqrt{1-\frac{{{c}^{2}}}{{{c}^{2}}}} \end{align}$ Simplify it further: $\begin{align} & \underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L={{L}_{0}}\cdot \sqrt{1-1} \\ & ={{L}_{0}}\cdot \sqrt{0} \\ & ={{L}_{0}}\cdot 0 \\ & =0 \end{align}$ Thus the limit $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L $ is 0. (b) From part (a), $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L=0$ This means that as the speed $ v $ of the starship approaches the speed of light, $ c $, the length of the starship $ L $ appears to approach 0 from the perspective of a stationary viewer on Earth. (c) The left hand limit is used in $\underset{v\to {{c}^{-}}}{\mathop{\lim }}\,L $ as it is not possible physically to exceed the speed of light $ c $. So, the velocity $ v $ by which the starship is moving can approach the speed of light $ c $ but must be less than $ c $.
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