## Calculus with Applications (10th Edition)

$$\frac{1}{{2\sqrt 5 }}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {5 + x} - \sqrt {5 - x} }}{{2x}} \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 {5 + x} - \sqrt {5 - x} } \right) = \sqrt {5 + 0} - \sqrt {5 - 0} = \sqrt 5 - \sqrt 5 = 0 \cr & \mathop {\lim }\limits_{x \to 0} 2x = 2\left( 0 \right) = 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 {5 + x} - \sqrt {5 - x} \to {D_x}\left( {\sqrt {5 + x} - \sqrt {5 - x} } \right) = \frac{1}{{2\sqrt {5 + x} }} - \frac{{ - 1}}{{2\sqrt {5 - x} }} \cr & {\text{for }}x \to {D_x}\left( {2x} \right) = 2 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {5 + x} - \sqrt {5 - x} }}{{2x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{2\sqrt {5 + x} }} + \frac{1}{{2\sqrt {5 - x} }}} \right) \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{2}\left( {\frac{1}{{2\sqrt {5 + 0} }} + \frac{1}{{2\sqrt {5 - 0} }}} \right) \cr & = \frac{1}{2}\left( {\frac{1}{{\sqrt 5 }}} \right) \cr & = \frac{1}{{2\sqrt 5 }} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {5 + x} - \sqrt {5 - x} }}{{2x}} = \frac{1}{{2\sqrt 5 }} \cr}