Question: \[\lim_{x\to -5}\frac{x^3+5x^2+x+5}{x^2+12x+35}\]
1. **State the problem:** We need to find the limit \(\lim_{x \to -5} \frac{x^3 + 5x^2 + x + 5}{x^2 + 12x + 35}\).
2. **Identify the form:** Substitute \(x = -5\) directly:
$$\frac{(-5)^3 + 5(-5)^2 + (-5) + 5}{(-5)^2 + 12(-5) + 35} = \frac{-125 + 125 - 5 + 5}{25 - 60 + 35} = \frac{0}{0}$$
This is an indeterminate form \(\frac{0}{0}\), so we need to simplify.
3. **Factor numerator and denominator:**
- Denominator: \(x^2 + 12x + 35 = (x + 5)(x + 7)\)
- Numerator: Factor by grouping:
$$x^3 + 5x^2 + x + 5 = (x^3 + 5x^2) + (x + 5) = x^2(x + 5) + 1(x + 5) = (x + 5)(x^2 + 1)$$
4. **Simplify the fraction:**
$$\frac{(x + 5)(x^2 + 1)}{(x + 5)(x + 7)}$$
Cancel the common factor \(x + 5\):
$$\frac{\cancel{(x + 5)}(x^2 + 1)}{\cancel{(x + 5)}(x + 7)} = \frac{x^2 + 1}{x + 7}$$
5. **Evaluate the limit:** Substitute \(x = -5\) into the simplified expression:
$$\frac{(-5)^2 + 1}{-5 + 7} = \frac{25 + 1}{2} = \frac{26}{2} = 13$$
**Final answer:** \(13\)