1. **Stating the problem:** Find the value of $\frac{a}{b}$ given that $$\lim_{x \to -1} \frac{x+3}{x^3 + ax + b} = +\infty.$$ 2. **Understanding the limit:** The limit tends to $+\infty$ as $x$ approaches $-1$. This means the denominator approaches zero from the positive side while the numerator is non-zero. 3. **Evaluate numerator at $x = -1$:** $$-1 + 3 = 2 \neq 0,$$ so the numerator is not zero at $x = -1$. 4. **Denominator must be zero at $x = -1$:** $$(-1)^3 + a(-1) + b = -1 - a + b = 0 \implies b = 1 + a.$$ 5. **Since the limit is $+\infty$, the denominator approaches zero positively:** Check the sign of the denominator near $x = -1$ by considering the derivative of the denominator: $$f(x) = x^3 + ax + b,$$ $$f'(-1) = 3(-1)^2 + a = 3 + a.$$ For the denominator to approach zero from the positive side as $x \to -1$, $f'(-1)$ must be positive: $$3 + a > 0 \implies a > -3.$$ 6. **Rewrite denominator as:** Since $x = -1$ is a root, $$x^3 + ax + b = (x + 1)(x^2 + px + q).$$ Expanding: $$x^3 + px^2 + qx + x^2 + px + q = x^3 + (p+1)x^2 + (q+p)x + q.$$ Matching coefficients with $x^3 + ax + b$: - Coefficient of $x^2$ is 0, so $p + 1 = 0 \implies p = -1$. - Coefficient of $x$ is $a$, so $q + p = a \implies q - 1 = a \implies q = a + 1$. - Constant term is $b$, so $q = b$. From step 4, $b = 1 + a$, so $q = b = 1 + a$. 7. **Check the quadratic factor:** $$x^2 - x + (a + 1).$$ 8. **For the limit to be $+\infty$, the denominator must change sign from negative to positive at $x = -1$.** Since the root at $x = -1$ is simple, the sign change depends on the quadratic factor at $x = -1$: $$(-1)^2 - (-1) + (a + 1) = 1 + 1 + a + 1 = a + 3.$$ For the denominator to be positive just to the right of $-1$, $a + 3 > 0 \implies a > -3$, consistent with step 5. 9. **Find $\frac{a}{b}$:** Recall from step 4: $$b = 1 + a,$$ so $$\frac{a}{b} = \frac{a}{1 + a}.$$ 10. **Use the condition that the denominator has a root at $x = -1$ and the limit tends to $+\infty$.** Try values from the options: - Option 1: $\frac{a}{b} = \frac{1}{2}$ $$a = \frac{1}{2}b,$$ $$b = 1 + a = 1 + \frac{1}{2}b,$$ $$b - \frac{1}{2}b = 1,$$ $$\frac{1}{2}b = 1 \implies b = 2,$$ $$a = \frac{1}{2} \times 2 = 1.$$ Check if $a = 1$ and $b = 2$ satisfy the conditions: - Denominator at $x = -1$: $$-1 + a + b = -1 + 1 + 2 = 2 \neq 0,$$ so no root at $x = -1$. - Option 2: $\frac{a}{b} = 2$ $$a = 2b,$$ $$b = 1 + a = 1 + 2b,$$ $$b - 2b = 1,$$ $$-b = 1 \implies b = -1,$$ $$a = 2 \times (-1) = -2.$$ Check denominator at $x = -1$: $$-1 - a + b = -1 - (-2) + (-1) = -1 + 2 - 1 = 0,$$ root confirmed. Check derivative: $$3 + a = 3 - 2 = 1 > 0,$$ so limit tends to $+\infty$. Therefore, the correct value is: $$\frac{a}{b} = 2.$$