Evaluate the limit: lim _{x rightarrow -9} fr | vedclass

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(D) Let $L = \lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$.Since the form is $\frac{0}{0}$,we apply $L$'Hopital's Rule:$L = \lim _{

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$-19 \log (0.4)$

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(D) Let $L = \lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$.Since the form is $\frac{0}{0}$,we apply $L$'Hopital's Rule:$L = \lim _{x \rightarrow -9} \frac{\frac{d}{dx}((2.5)^{81-x^2}) - \frac{d}{dx}((0.4)^{x+9})}{\frac{d}{dx}(x+9)}$$L = \lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2} \cdot \ln(2.5) \cdot (-2x) - (0.4)^{x+9} \cdot \ln(0.4) \cdot 1}{1}$Substitute $x = -9$:$L = (2.5)^{81-81} \cdot \ln(2.5) \cdot (-2(-9)) - (0.4)^{0} \cdot \ln(0.4)$$L = (2.5)^0 \cdot \ln(2.5) \cdot 18 - 1 \cdot \ln(0.4)$$L = 18 \ln(2.5) - \ln(0.4)$.Since $\ln(0.4) = \ln(2.5^{-1}) = -\ln(2.5)$,we have $L = 18 \ln(2.5) + \ln(2.5) = 19 \ln(2.5)$.However,checking the options provided,the expression $-19 \log(0.4)$ is equivalent to $-19 \ln(0.4) / \ln(10) = -19 \ln(2.5^{-1}) / \ln(10) = 19 \ln(2.5) / \ln(10) = 19 \log(2.5)$.Given the structure,option $D$ is the intended answer.

Evaluate the limit: $\lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$

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