Search any question & find its solution
Question:
Answered & Verified by Expert
$$
\lim _{x \rightarrow 9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}=
$$
Options:
\lim _{x \rightarrow 9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}=
$$
Solution:
2917 Upvotes
Verified Answer
The correct answer is:
$-19 \log (0.4)$
$\begin{aligned} & \lim _{2 \rightarrow-9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9} \\ = & \lim _{x \rightarrow-9} \frac{(2.5)^{81}(-2 x)(2.5)^{-x^2} \ln (2.5)-(0.4)^{(x+9)} \log (0.4)}{1} \\ = & 1 .(18) \log (2.5)-1 . \log (0.4)=-19 \log (0.4)\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.