Question: 1) $\lim_{x\to -5}\frac{x^3+5x^2+x+5}{x^2+12x+35}$
2) Graph/shape: none visible; math expression is centered. position_hint = center
1. **State the problem:**
Find the limit
$$\lim_{x\to -5}\frac{x^3+5x^2+x+5}{x^2+12x+35}$$
2. **Recall the formula and rules:**
To find limits involving rational functions, if direct substitution leads to an indeterminate form like $\frac{0}{0}$, we try to factor numerator and denominator and simplify.
3. **Evaluate direct substitution:**
Substitute $x = -5$:
Numerator: $(-5)^3 + 5(-5)^2 + (-5) + 5 = -125 + 125 - 5 + 5 = 0$
Denominator: $(-5)^2 + 12(-5) + 35 = 25 - 60 + 35 = 0$
So direct substitution gives $\frac{0}{0}$, an indeterminate form.
4. **Factor numerator and denominator:**
Denominator:
$$x^2 + 12x + 35 = (x + 5)(x + 7)$$
Numerator:
Group terms:
$$x^3 + 5x^2 + x + 5 = (x^3 + 5x^2) + (x + 5) = x^2(x + 5) + 1(x + 5) = (x + 5)(x^2 + 1)$$
5. **Simplify the expression:**
$$\frac{(x + 5)(x^2 + 1)}{(x + 5)(x + 7)}$$
Cancel the common factor $x + 5$:
$$\frac{\cancel{(x + 5)}(x^2 + 1)}{\cancel{(x + 5)}(x + 7)} = \frac{x^2 + 1}{x + 7}$$
6. **Evaluate the limit of the simplified expression:**
Substitute $x = -5$:
$$\frac{(-5)^2 + 1}{-5 + 7} = \frac{25 + 1}{2} = \frac{26}{2} = 13$$
**Final answer:**
$$\lim_{x\to -5}\frac{x^3+5x^2+x+5}{x^2+12x+35} = 13$$