To solve the limit \[ \lim_{x \to 2} \frac{x^3 - x - 6}{x^4 - 2x - 12}, \] we first plug in \( x = 2 \): 1. For the numerator: \[ 2^3 - 2 - 6 = 8 - 2 - 6 = 0. \] 2. For the denominator: \[ 2^4 - 2(2) - 12 = 16 - 4 - 12 = 0. \] Since both the numerator and denominator equal zero when \( x = 2 \), we have an indeterminate form \( \frac{0}{0} \). We can apply techniques such as factoring or L'Hôpital's rule. ### Step 1: Factor the numerator First, let's try to factor the numerator \( x^3 - x - 6 \). We can use synthetic division or polynomial division to see if \( x - 2 \) is a factor (because we suspect \( x = 2 \) is a root). Using synthetic division: \[ \begin{array}{r|rrrr} 2 & 1 & 0 & -1 & -6 \\ & & 2 & 4 & 6 \\ \hline & 1 & 2 & 3 & 0 \\ \end{array} \] This gives us \( 1x^2 + 2x + 3 \), meaning: \[ x^3 - x - 6 = (x - 2)(x^2 + 2x + 3). \] ### Step 2: Factor the denominator Next, we will factor the denominator \( x^4 - 2x - 12 \). We also check if \( x - 2 \) is a factor using synthetic division: \[ \begin{array}{r|rrrrr} 2 & 1 & 0 & 0 & -2 & -12 \\ & & 2 & 4 & 8 & 12 \\ \hline & 1 & 2 & 4 & 6 & 0 \\ \end{array} \] This gives us \( 1x^3 + 2x^2 + 4x + 6 \), meaning: \[ x^4 - 2x - 12 = (x - 2)(x^3 + 2x^2 + 4x + 6). \] ### Step 3: Simplifying the limit expression Now, we can rewrite the limit expression: \[ \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 3)}{(x - 2)(x^3 + 2x^2 + 4x + 6)}. \] Cancel the common factor \( x - 2 \) (valid because we are taking the limit as \( x \) approaches 2, not at \( x = 2 \)): \[ \lim_{x \to 2} \frac{x^2 + 2x + 3}{x^3 + 2x^2 + 4x + 6}. \] ### Step 4: Evaluate at \( x = 2 \) Now we can substitute \( x = 2 \): 1. For the numerator: \[ 2^2 + 2(2) + 3 = 4 + 4 + 3 = 11. \] 2. For the denominator: \[ 2^3 + 2(2^2) + 4(2) + 6 = 8 + 8 + 8 + 6 = 30. \] Thus, we get: \[ \frac{11}{30}. \] ### Conclusion Thus, the value of the limit is: \[ \lim_{x \to 2} \frac{x^3 - x - 6}{x^4 - 2x - 12} = \frac{11}{30}. \]