Precalculus (6th Edition) Blitzer

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.
(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 $c$.