1. Problem statement: Find the limit as $x$ approaches $-5$ of the function. $$\lim_{x\to -5} \frac{\frac{1}{5}+\frac{1}{x}}{10+2x}$$ 2. Direct substitution test: Substitute $x=-5$ into numerator and denominator to check for an indeterminate form. Numerator: $\frac{1}{5}+\frac{1}{-5}=\frac{1}{5}-\frac{1}{5}=0$. Denominator: $10+2(-5)=10-10=0$. 3. Method: Use algebraic simplification by combining fractions and factoring to remove the common factor that causes the $0/0$ indeterminate form. 4. Combine the numerator fractions: $$\frac{1}{5}+\frac{1}{x}=\frac{x+5}{5x}$$ 5. Rewrite the whole expression and simplify: $$\frac{\frac{x+5}{5x}}{10+2x}=\frac{x+5}{5x(10+2x)}$$ 6. Factor the denominator and cancel the common factor $x+5$ (valid for $x\ne -5$): $$10+2x=2(x+5)$$ Hence $$\frac{x+5}{5x(10+2x)}=\frac{x+5}{5x\cdot 2(x+5)}=\frac{1}{10x}\quad (x\ne -5)$$ 7. Evaluate the limit of the simplified expression: $$\lim_{x\to -5}\frac{1}{10x}=\frac{1}{10(-5)}=-\frac{1}{50}$$ 8. Final answer: The limit is $-\frac{1}{50}$.