## University Calculus: Early Transcendentals (3rd Edition)

- As $v$ increases, $L$ will decrease. - $\lim_{v\to c^-}L=0$ - The left-hand limit was needed because the speed $v$ of the object cannot exceed the speed of light in a vacuum.
$$L=L_0\sqrt{1-\frac{v^2}{c^2}}$$ As $v$ increases, $\frac{v^2}{c^2}$ will also increase, and $1-\frac{v^2}{c^2}$, and $\sqrt{1-\frac{v^2}{c^2}}$ following, as a result, will decrease. The final result is that $L$, or the length of the object, will decrease. $$\lim_{v\to c^-}L=\lim_{v\to c^-}\Big(L_0\sqrt{1-\frac{v^2}{c^2}}\Big)$$ $$\lim_{v\to c^-}L=L_0\sqrt{1-\frac{c^2}{c^2}}=L_0\sqrt{1-1}=L_0\times0=0$$ The left-hand limit was needed because $v\lt c$, or the speed of the object must always be smaller than the speed of light in a vacuum. Material objects cannot exceed the speed of light. In fact, if you examine the formula of the Lorentz contraction, the closer the speed $v$ of the objects is to that of light, the more contracted the object appears in the eye of the observer. And if $v=c$, $L=0$, and the object basically disappears from the observer. If $v\gt c$, the values of $L\lt0$, or the contracted length is now negative, which is impossible.
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