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

Question

Aicharya · Accepted Answer

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

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

Solution