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\( \operatorname { Li } _ { x \rightarrow - 3 } \frac { x ^ { 3 } - 17 x - 30 } { x ^ { 2 } - 2 x - 15 } \)

Ask by Fuentes Summers. in Nigeria
Feb 11,2025

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The limit does not exist because the denominator becomes zero when \( x \) approaches \(-3\), leading to division by zero.

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To evaluate the limit as \( x \) approaches -3 for the expression \( \frac{x^3 - 17x - 30}{x^2 - 2x - 15} \), we start by substituting \( x = -3 \): 1. **Numerator:** \[ (-3)^3 - 17(-3) - 30 = -27 + 51 - 30 = -6 \] 2. **Denominator:** \[ (-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0 \] Since the denominator is 0 and the numerator is -6, this means we initially have an indeterminate form. Therefore, we need to factor the expression to simplify. 3. **Factoring the denominator:** \[ x^2 - 2x - 15 = (x - 5)(x + 3) \] 4. **Factoring the numerator:** We will try to perform synthetic division or polynomial long division for \( x^3 - 17x - 30 \) by \( x + 3 \). Dividing \( x^3 - 17x - 30 \) by \( x + 3 \): - The first term yields \( x^2 \). - \( x^2(x + 3) = x^3 + 3x^2 \) and therefore subtract: \[ (x^3 - 17x - 30) - (x^3 + 3x^2) = -3x^2 - 17x - 30 \] - Now take \( -3x \): \[ -3x(x + 3) = -3x^2 - 9x \] So, \[ (-3x^2 - 17x - 30) - (-3x^2 - 9x) = -8x - 30 \] - Finally, take \( -8 \): \[ -8(x + 3) = -8x - 24 \] Combine it all: \[ (-8x - 30) - (-8x - 24) = -6 \] Thus, we get: \[ x^3 - 17x - 30 = (x + 3)(x^2 - 3x - 10) \] So, we have: \[ \frac{x^3 - 17x - 30}{x^2 - 2x - 15} = \frac{(x + 3)(x^2 - 3x - 10)}{(x - 5)(x + 3)} \] Cancel the \( (x + 3) \): \[ \frac{x^2 - 3x - 10}{x - 5} \] Now we can substitute \( x = -3 \): Numerator: \[ (-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 \] Denominator: \[ -3 - 5 = -8 \] Finally: \[ \operatorname{Li}_{x \to -3} \frac{x^3 - 17x - 30}{x^2 - 2x - 15} = \frac{8}{-8} = -1 \] Thus, the limit is: \[ \boxed{-1} \]

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