Let's solve the limit step by step: \[ \lim_{x \to 2} \frac{16 - x^4}{x^3 - 8} \] ### **Step 1: Substitute \( x = 2 \)** \[ \frac{16 - (2)^4}{(2)^3 - 8} = \frac{16 - 16}{8 - 8} = \frac{0}{0} \] This is an **indeterminate form**, so we need to factor and simplify. --- ### **Step 2: Factor Numerator and Denominator** #### **Numerator: \( 16 - x^4 \)** This is a difference of squares: \[ 16 - x^4 = (4^2 - x^2)(4^2 + x^2) = (4-x^2)(4+x^2) \] But \(4-x^2\) is also a difference of squares: \[ 4 - x^2 = (2-x)(2+x) \] So, \[ 16 - x^4 = (2-x)(2+x)(4+x^2) \] #### **Denominator: \( x^3 - 8 \)** This is a difference of cubes: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \] --- ### **Step 3: Substitute Factored Forms** \[ \lim_{x \to 2} \frac{(2-x)(2+x)(4+x^2)}{(x-2)(x^2 + 2x + 4)} \] Notice \(2-x\) and \(x-2\) are negatives of each other: \[ 2-x = -(x-2) \] So, \[ \frac{(2-x)}{(x-2)} = -1 \] --- ### **Step 4: Simplify** \[ \lim_{x \to 2} \frac{-(2+x)(4+x^2)}{x^2 + 2x + 4} \] --- ### **Step 5: Substitute \(x = 2\) again** \[ -(2+2)(4+2^2) / (2^2 + 2 \cdot 2 + 4) \] \[ = - (4)(4+4) / (4 + 4 + 4) \] \[ = - (4)(8) / 12 \] \[ = -32 / 12 \] \[ = -8/3 \] --- ### **Final Answer** \[ \boxed{-\frac{8}{3}} \]

Let's carefully evaluate the limit: \[ \lim_{x \to 2} \frac{16 - x^4}{x^3 - 8} \] --- ### **Step 1: Check the direct substitution** Substitute \(x = 2\): Numerator: \[ 16 - (2)^4 = 16 - 16 = 0 \] Denominator: \[ (2)^3 - 8 = 8 - 8 = 0 \] We get a \(\frac{0}{0}\) indeterminate form, so we need to simplify further. --- ### **Step 2: Factor numerator and denominator** **Numerator:** \(16 - x^4\) - Recognize this as a difference of squares: \[ 16 - x^4 = (4)^2 - (x^2)^2 = (4 - x^2)(4 + x^2) \] - Further factor \(4 - x^2\) as a difference of squares: \[ 4 - x^2 = (2 - x)(2 + x) \] Thus, numerator: \[ 16 - x^4 = (2 - x)(2 + x)(4 + x^2) \] --- **Denominator:** \(x^3 - 8\) - Recognize as a difference of cubes: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \] --- ### **Step 3: Rewrite the limit with factored forms** \[ \lim_{x \to 2} \frac{(2 - x)(2 + x)(4 + x^2)}{(x - 2)(x^2 + 2x + 4)} \] Notice that \(2 - x\) and \(x - 2\) are negatives of each other: \[ 2 - x = -(x - 2) \] So, \[ \frac{(2 - x)}{(x - 2)} = -1 \] --- ### **Step 4: Simplify the expression** The limit becomes: \[ \lim_{x \to 2} \frac{-(x - 2)(2 + x)(4 + x^2)}{(x - 2)(x^2 + 2x + 4)} \] Cancel \((x - 2)\): \[ \lim_{x \to 2} \frac{-(2 + x)(4 + x^2)}{x^2 + 2x + 4} \] --- ### **Step 5: Substitute \(x = 2\)** Calculate numerator: \[ -(2 + 2)(4 + 2^2) = - (4)(4 + 4) = - (4)(8) = -32 \] Calculate denominator: \[ (2)^2 + 2 \cdot 2 + 4 = 4 + 4 + 4 = 12 \] --- ### **Step 6: Final calculation** \[ \frac{-32}{12} = -\frac{8}{3} \] --- ## **Final Answer:** \[ \boxed{-\frac{8}{3}} \] This is the value of the limit.

Let's carefully evaluate the limit: \[ \lim_{x \to 2} \frac{16 - x^4}{x^3 - 8} \] ### Step 1: Check direct substitution Substitute \(x = 2\): - Numerator: \(16 - (2)^4 = 16 - 16 = 0\) - Denominator: \(8 - 8 = 0\) We get an indeterminate form \(\frac{0}{0}\), so we need to simplify. --- ### Step 2: Factor numerator and denominator **Numerator:** \(16 - x^4\) - Recognize as a difference of squares: \[ 16 - x^4 = (4)^2 - (x^2)^2 = (4 - x^2)(4 + x^2) \] - Factor \(4 - x^2\) further as a difference of squares: \[ 4 - x^2 = (2 - x)(2 + x) \] Thus, \[ 16 - x^4 = (2 - x)(2 + x)(4 + x^2) \] **Denominator:** \(x^3 - 8\) - Recognize as a difference of cubes: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \] --- ### Step 3: Rewrite the limit with factored forms \[ \lim_{x \to 2} \frac{(2 - x)(2 + x)(4 + x^2)}{(x - 2)(x^2 + 2x + 4)} \] Note that: \[ 2 - x = -(x - 2) \] Therefore, \[ \frac{(2 - x)}{(x - 2)} = -1 \] --- ### Step 4: Simplify the expression Replace \((2 - x)\) with \(- (x - 2)\): \[ \lim_{x \to 2} \frac{- (x - 2)(2 + x)(4 + x^2)}{(x - 2)(x^2 + 2x + 4)} \] Cancel \((x - 2)\) from numerator and denominator: \[ \lim_{x \to 2} \frac{ - (2 + x)(4 + x^2)}{x^2 + 2x + 4} \] ### Step 5: Substitute \(x = 2\) Calculate numerator: \[ -(2 + 2)(4 + 2^2) = - (4)(4 + 4) = - (4)(8) = -32 \] Calculate denominator: \[ 2^2 + 2 \times 2 + 4 = 4 + 4 + 4 = 12 \] ### Step 6: Final value \[ \frac{-32}{12} = -\frac{8}{3} \] --- ## **Final answer:** \[ \boxed{-\frac{8}{3}} \] This is the value of the limit.