Q99 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

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

  • A
    $18 \log (2.5)+\log (0.4)$
  • B
    $\log (2.5)-\log (0.4)$
  • C
    $18(\log (2.5)+\log (0.4))$
  • Correct answer
    D
    $-19 \log (0.4)$

Solution

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

Hence, the correct answer is option: 'D'.

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