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AP EAMCET · Maths · Limits

\[
\lim _{x \rightarrow 9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}=
\]

  1. A \(18 \log (2.5)+\log (0.4)\)
  2. B \(\log (2.5)-\log (0.4)\)
  3. C \(18(\log (2.5)+\log (0.4))\)
  4. D \(-19 \log (0.4)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-19 \log (0.4)\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \lim _{2 \rightarrow-9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9} \\ = & \lim _{x \rightarrow-9} \frac{(2.5)^{81}(-2 x)(2.5)^{-x^2} \ln (2.5)-(0.4)^{(x+9)} \log (0.4)}{1} \\ = & 1 .(18) \log (2.5)-1 . \log (0.4)=-19 \log (0.4)\end{aligned}\)