1. **State the problem:** Find the limit as $x$ approaches $-5$ of the expression $$\frac{\frac{1}{5} + \frac{1}{x}}{10 + 2x}.$$\n\n2. **Rewrite the expression:** The expression can be written as $$\frac{\frac{1}{5} + \frac{1}{x}}{10 + 2x} = \frac{\frac{1}{5} + \frac{1}{x}}{10 + 2x}.$$\n\n3. **Find a common denominator in the numerator:**\n$$\frac{1}{5} + \frac{1}{x} = \frac{x}{5x} + \frac{5}{5x} = \frac{x + 5}{5x}.$$\n\n4. **Substitute back:**\n$$\frac{\frac{x + 5}{5x}}{10 + 2x} = \frac{x + 5}{5x(10 + 2x)}.$$\n\n5. **Evaluate the limit as $x \to -5$:**\nSubstitute $x = -5$:\n$$\frac{-5 + 5}{5 \cdot (-5) \cdot (10 + 2 \cdot (-5))} = \frac{0}{5 \cdot (-5) \cdot (10 - 10)} = \frac{0}{5 \cdot (-5) \cdot 0} = \frac{0}{0}.$$\nThis is an indeterminate form, so we need to simplify further.\n\n6. **Factor numerator and denominator:**\nNotice numerator has factor $(x + 5)$, denominator has factor $(10 + 2x) = 2(x + 5)$. So:\n$$\frac{x + 5}{5x(10 + 2x)} = \frac{x + 5}{5x \cdot 2(x + 5)} = \frac{1}{5x \cdot 2} = \frac{1}{10x}, \quad x \neq -5.$$\n\n7. **Now evaluate the simplified expression at $x = -5$:**\n$$\lim_{x \to -5} \frac{1}{10x} = \frac{1}{10 \cdot (-5)} = \frac{1}{-50} = -\frac{1}{50}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{50}}.$$