1. **State the problem:** Find the limit $ \lim_{x \to -1} \frac{x^2 + x}{(x+1)^4} $. 2. **Recall the formula and rules:** When evaluating limits involving rational functions, if direct substitution leads to an indeterminate form like $ \frac{0}{0} $, factor and simplify the expression. 3. **Factor numerator:** $ x^2 + x = x(x+1) $. 4. **Rewrite the limit:** $$\lim_{x \to -1} \frac{x(x+1)}{(x+1)^4} = \lim_{x \to -1} \frac{x \cancel{(x+1)}}{\cancel{(x+1)} (x+1)^3} = \lim_{x \to -1} \frac{x}{(x+1)^3}$$ 5. **Evaluate the simplified limit:** Substitute $ x = -1 $: $$\frac{-1}{(0)^3} = \frac{-1}{0}$$ 6. **Interpretation:** Division by zero indicates the limit tends to infinity or negative infinity depending on the sign of the denominator near $ x = -1 $. 7. **Check the sign of denominator near $ x = -1 $:** - For $ x \to -1^+ $, $ (x+1)^3 > 0 $, so the fraction $ \frac{x}{(x+1)^3} \to -\infty $. - For $ x \to -1^- $, $ (x+1)^3 < 0 $, so the fraction $ \frac{x}{(x+1)^3} \to +\infty $. **Final answer:** The limit does not exist because the left and right limits are not equal: $$\lim_{x \to -1^-} \frac{x^2 + x}{(x+1)^4} = +\infty, \quad \lim_{x \to -1^+} \frac{x^2 + x}{(x+1)^4} = -\infty$$