8. \( \lim _{x \rightarrow-} \frac{\sqrt{x+2}-\sqrt{2}}{x} \)

To solve the limit \( \lim_{x \rightarrow 0^-} \frac{\sqrt{x+2} - \sqrt{2}}{x} \), we can start by substituting \( x = 0 \) to see if the limit can be evaluated directly: \[ \frac{\sqrt{0+2} - \sqrt{2}}{0} = \frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0} \] Since this results in an indeterminate form, we can use L'Hôpital's Rule. First, we compute the derivative of the numerator and the denominator. The numerator is \( \sqrt{x+2} - \sqrt{2} \): 1. The derivative of \( \sqrt{x+2} \) is \( \frac{1}{2\sqrt{x+2}} \). 2. The derivative of \( \sqrt{2} \) is 0. Thus, the derivative of the numerator is \( \frac{1}{2\sqrt{x+2}} \). The denominator is \( x \): 1. The derivative of \( x \) is 1. Using L'Hôpital's Rule, we now have: \[ \lim_{x \rightarrow 0^-} \frac{\sqrt{x+2} - \sqrt{2}}{x} = \lim_{x \rightarrow 0^-} \frac{\frac{1}{2\sqrt{x+2}}}{1} = \frac{1}{2\sqrt{0+2}} = \frac{1}{2\sqrt{2}} \] So the limit evaluates to: \[ \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \] Therefore, the final answer is: \[ \frac{\sqrt{2}}{4} \]