lim x->-7 (sqrt(2-x)-3)/(x+7)

1. Identify the limit: The problem asks to evaluate the limit $\lim_{x \to -7} \frac{\sqrt{2-x}-3}{x+7}$.

1. Direct substitution: If we substitute $x = -7$ directly, we get $\frac{\sqrt{2-(-7)}-3}{-7+7} = \frac{\sqrt{9}-3}{0} = \frac{0}{0}$. This is an indeterminate form, suggesting we should simplify the expression.

3. Rationalize the numerator: To simplify, multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{2-x}+3$:
$$\lim_{x \to -7} \frac{(\sqrt{2-x}-3)(\sqrt{2-x}+3)}{(x+7)(\sqrt{2-x}+3)}$$

4. Simplify the expression: Using the difference of squares formula $(a-b)(a+b) = a^2-b^2$:
$$\lim_{x \to -7} \frac{(2-x) - 9}{(x+7)(\sqrt{2-x}+3)} = \lim_{x \to -7} \frac{-x-7}{(x+7)(\sqrt{2-x}+3)}$$

5. Cancel common factors: Factor out $-1$ from the numerator:
$$\lim_{x \to -7} \frac{-(x+7)}{(x+7)(\sqrt{2-x}+3)} = \lim_{x \to -7} \frac{-1}{\sqrt{2-x}+3}$$

6. Evaluate the limit: Now substitute $x = -7$ into the simplified expression:
$$\frac{-1}{\sqrt{2-(-7)}+3} = \frac{-1}{\sqrt{9}+3} = \frac{-1}{3+3} = -\frac{1}{6}$$

1. Convert to decimal: The value $-\frac{1}{6} \approx -0.166666...$, which rounds to $-0.1667$ as requested.