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 $.
