Question

The value of $\underset{x\to 0}{\lim }\frac{9ln\left(2-cos25x\right)}{5{\left(ln\right)}^2\left(sin3x+1\right)}$ is equal to

Answer 1

Answer: 62.5

$\underset{x\to 0}{\lim }\frac{9ln\left(2-cos25x\right)}{5{\left(ln\right)}^2\left(sin3x+1\right)}=\underset{x\to 0}{\lim }\frac{9ln\left\{1+\left(1-cos25x\right)\right\}}{5{\left(ln\right)}^2\left(1+sin3x\right)}$
$=\underset{x\to 0}{\lim }\frac{9\left(1-cos25x\right)}{5{\left(sin3x\right)}^2}$
$\Rightarrow \underset{x\to 0}{\lim }\frac{9{\left(25x\right)}^2}{10{\left(3x\right)}^2}=\frac{625}{10}=62.5$