## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 104: 56

#### Answer

$\lim\limits_{v \to c^-}L=0$ As an object moves faster and very close to the speed of life, its length contracts, becoming closer to 0. The left hand limit is necessary as an object can not travel faster than the speed of light, so $v$ can only be less than or equal to $c$. Also, $1-\frac{v^2}{c^2}\geq0, v^2\leq c^2, v\leq c$ as the direction the object is travelling in does not matter here.

#### Work Step by Step

$\lim\limits_{v \to c^-}L=\lim\limits_{v \to c^-}L_0\sqrt{1-v^2/c^2}$ $=L_0\sqrt{1-(c^-)^2/c^2}$ $=L_0\sqrt{1-(c^2)^-/c^2}$ $=L_0\sqrt {1-1^-}$ $=L_0\sqrt{0^+}$ $=0$ As an object moves faster and very close to the speed of life, its length contracts, becoming closer to 0. The left hand limit is necessary as an object can not travel faster than the speed of light, so $v$ can only be less than or equal to $c$. Also, $1-\frac{v^2}{c^2}\geq0, v^2\leq c^2, v\leq c$ as the direction the object is travelling in does not matter here.

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