- For problem 4, direct substitution of $x=1$ results in the indeterminate form $\frac{0}{0}$. Polynomial division or factoring is required. Factoring the numerator $4x^3 - 2x^2 + x - 3$ by $(x-1)$ yields $(x-1)(4x^2 + 2x + 3)$. The limit simplifies to $\lim_{x \to 1} (4x^2 + 2x + 3) = 4(1)^2 + 2(1) + 3 = 9$.
- For problem 5, direct substitution of $x=4$ gives $\frac{0}{0}$. Factor both the numerator and denominator. The numerator $2x^2 - 12x + 16$ factors as $2(x-2)(x-4)$. The denominator $x^2 - x - 12$ factors as $(x-4)(x+3)$. Canceling the common factor $(x-4)$ leaves $\lim_{x \to 4} \frac{2(x-2)}{x+3} = \frac{2(4-2)}{4+3} = \frac{4}{7}$.
- For problem 6, direct substitution of $x=-3$ gives $\frac{0}{0}$. Factor the numerator $x^2 + 2x - 3$ as $(x+3)(x-1)$. Factor the denominator $-3x - 9$ as $-3(x+3)$. Canceling the common factor $(x+3)$ leaves $\lim_{x \to -3} \frac{x-1}{-3} = \frac{-3-1}{-3} = \frac{-4}{-3} = \frac{4}{3}$.
Parent Tip: Review the logic above to help your child master the concept of limits worksheet.
