Question

The value of $\displaystyle\lim _{x \rightarrow 0} \frac{1}{x^{18}}\left(1-\cos \left(\frac{x^3}{3}\right)-\cos \left(\frac{x^6}{6}\right)+\cos \left(\frac{x^3}{3}\right) \cdot \cos \left(\frac{x^6}{6}\right)\right)$ is $\lambda^2$, then the value of $900 \lambda$ is equal to (here, $\lambda>0$ )

Answer 1

$\displaystyle\lim _{x \rightarrow 0}\left\{1-\cos \left(\frac{x^3}{3}\right)-\cos \left(\frac{x^6}{6}\right)\left(1-\cos \left(\frac{x^3}{3}\right)\right)\right\} \cdot \frac{1}{x^{18}}$
$\displaystyle\lim _{x \rightarrow 0} \frac{\left(1-\cos \left(\frac{x^3}{3}\right)\right)\left(1-\cos \left(\frac{x^6}{6}\right)\right)}{x^{18}}$
$\displaystyle\lim _{x \rightarrow 0} \frac{1-\cos \left(\frac{x^3}{3}\right)}{\left(\frac{x^3}{3}\right)^2}\left(\frac{x^6}{9}\right) \cdot \frac{1-\cos \left(\frac{x^6}{6}\right)}{\left(\frac{x^6}{6}\right)^2} \cdot\left(\frac{x^6}{6}\right)^2 \cdot \frac{1}{x^{18}}$
$\displaystyle\lim _{x \rightarrow 0} \frac{1}{2} \times \frac{x^6}{9} \cdot \frac{1}{2} \cdot \frac{x^{12}}{36} \cdot \frac{1}{x^{18}}=\frac{1}{(36)^2}$
$\therefore \lambda=\frac{1}{36}$