Answer
$$ - \frac{1}{{\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 - x} - \sqrt {3 + x} }}{x} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {3 - x} - \sqrt {3 + x} } \right) = \sqrt {3 - 0} - \sqrt {3 + 0} = \sqrt 3 - \sqrt 3 = 0 \cr
& \mathop {\lim }\limits_{x \to 0} x = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{.}} \cr
& {\text{for }}\sqrt {3 - x} - \sqrt {3 + x} \to {D_x}\left( {\sqrt {3 - x} - \sqrt {3 + x} } \right) = \frac{{ - 1}}{{2\sqrt {3 - x} }} - \frac{1}{{2\sqrt {3 + x} }} \cr
& {\text{for }}x \to {D_x}\left( x \right) = 1 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 - x} - \sqrt {3 + x} }}{x} = \mathop {\lim }\limits_{x \to 0} \left( { - \frac{1}{{2\sqrt {3 - x} }} - \frac{1}{{2\sqrt {3 + x} }}} \right) \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = - \frac{1}{{2\sqrt {3 - 0} }} - \frac{1}{{2\sqrt {3 + 0} }} \cr
& = - \frac{1}{{\sqrt 3 }} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 - x} - \sqrt {3 + x} }}{x} = - \frac{1}{{\sqrt 3 }} \cr} $$
