AP EAMCET · Maths · Limits
\[
\lim _{x \rightarrow 9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}=
\]
- A \(18 \log (2.5)+\log (0.4)\)
- B \(\log (2.5)-\log (0.4)\)
- C \(18(\log (2.5)+\log (0.4))\)
- D \(-19 \log (0.4)\)
Answer & Solution
Correct Answer
(D) \(-19 \log (0.4)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{2 \rightarrow-9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9} \\ = & \lim _{x \rightarrow-9} \frac{(2.5)^{81}(-2 x)(2.5)^{-x^2} \ln (2.5)-(0.4)^{(x+9)} \log (0.4)}{1} \\ = & 1 .(18) \log (2.5)-1 . \log (0.4)=-19 \log (0.4)\end{aligned}\)
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