Question

The value of $\underset{x \rightarrow 0}{lim} \frac{9 ln ⁡ \left(2 - cos ⁡ 25 x\right)}{5 \left(ln\right)^{2} \left(sin ⁡ 3 x + 1\right) ⁡}$ is equal to

Answer 1

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