1. **State the problem:** Find the limit as $x$ approaches $-5$ of the expression $$\frac{\frac{1}{5+x}}{10+2x}.$$\n\n2. **Rewrite the expression:** The complex fraction can be simplified by dividing the numerator by the denominator:\n$$\frac{\frac{1}{5+x}}{10+2x} = \frac{1}{5+x} \times \frac{1}{10+2x} = \frac{1}{(5+x)(10+2x)}.$$\n\n3. **Substitute $x = -5$ directly:**\n$$ (5 + (-5)) = 0, \quad (10 + 2(-5)) = 10 - 10 = 0.$$\nSo the denominator becomes $0 \times 0 = 0$, which is undefined. We have an indeterminate form $\frac{1}{0}$ which suggests the limit might be infinite or does not exist.\n\n4. **Analyze the behavior near $x = -5$:**\nLet $x = -5 + h$ where $h \to 0$. Then:\n$$ (5 + x) = h, \quad (10 + 2x) = 10 + 2(-5 + h) = 10 - 10 + 2h = 2h.$$\nSo the expression becomes:\n$$ \frac{1}{h \times 2h} = \frac{1}{2h^2}.$$\n\n5. **Evaluate the limit as $h \to 0$:**\nSince $h^2$ is always positive and approaches $0$, $\frac{1}{2h^2} \to +\infty$.\n\n**Final answer:**\n$$\lim_{x \to -5} \frac{\frac{1}{5+x}}{10+2x} = +\infty.$$