Question

\( \lim_{x \to -9} \frac{(2.5)^{81 - x^2} - (0.4)^{x + 9}}{x + 9} = \)

• A. \( 18 \log(2.5) + \log(0.4) \)
• B. \( \log (2.5) - \log (0.4) \)
• C. \( 18(\log(2.5) + \log(0.4)) \)
• D. \( -19 \log(0.4) \)

Hint:

Double-check the problem statement and options for any potential errors when the derived solution does not match the provided answer.

Answer 1

The Correct Option is D

Step 1: Express all terms with the same base. We know: \[ (0.4)^{x + 9} = \left( \frac{2}{5} \right)^{x + 9} = \left( \frac{1}{2.5} \right)^{x + 9} = (2.5)^{-(x + 9)} \] So the given expression becomes: \[ \lim_{x \to -9} \frac{(2.5)^{81 - x^2} - (2.5)^{-(x + 9)}}{x + 9} \]
Step 2: Apply L'Hôpital's Rule since it is an indeterminate form \( \frac{0}{0} \). Differentiate the numerator and denominator: \[ \frac{d}{dx}\left( (2.5)^{81 - x^2} \right) = (2.5)^{81 - x^2} \cdot \ln(2.5) \cdot (-2x) \] \[ \frac{d}{dx}\left( (2.5)^{-(x + 9)} \right) = (2.5)^{-(x + 9)} \cdot (-\ln(2.5)) \] So derivative of numerator is: \[ -2x \cdot (2.5)^{81 - x^2} \cdot \ln(2.5) + (2.5)^{-(x + 9)} \cdot \ln(2.5) \] And derivative of denominator is: \[ 1 \] Now substitute \( x = -9 \): \[ -2x = 18,\quad 81 - x^2 = 0 \Rightarrow (2.5)^0 = 1,\quad (2.5)^{-(x + 9)} = (2.5)^0 = 1 \] Thus: \[ \lim_{x \to -9} = 18 \cdot 1 \cdot \ln(2.5) + 1 \cdot \ln(2.5) = 19 \ln(2.5) \]
Step 3: Convert to desired base. We know: \[ \ln(0.4) = \ln\left( \frac{2}{5} \right) = -\ln(2.5) \Rightarrow \ln(2.5) = -\ln(0.4) \] Hence: \[ 19 \ln(2.5) = 19 \cdot (-\ln(0.4)) = -19 \ln(0.4) \]