1. **State the problem:** Find the limit $$\lim_{x \to 9^-} \frac{1+x}{x-9}$$. 2. **Recall the limit concept:** We want to see what value the function approaches as $x$ gets closer to 9 from the left side (values less than 9). 3. **Analyze the numerator and denominator:** - Numerator: $1+x$. As $x \to 9$, numerator approaches $1+9=10$. - Denominator: $x-9$. As $x \to 9^-$, $x$ is slightly less than 9, so $x-9$ is a small negative number approaching 0 from the negative side. 4. **Evaluate the limit:** Since numerator approaches 10 (a positive number) and denominator approaches 0 from the negative side, the fraction behaves like $\frac{10}{\text{small negative}}$ which tends to $-\infty$. 5. **Final answer:** $$\lim_{x \to 9^-} \frac{1+x}{x-9} = -\infty$$