Limiting Cosines of Singular Double Triple

To evaluate this limit,we will use the expansions of $\cos(x) , \cos(2x) \text{ \& } \cos(3x)$

$\cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \ldots$

$\cos(2x) = 1 - \dfrac{4x^2}{2!} + \dfrac{16x^3}{4!} \ldots$

$\cos(3x) = 1 - \dfrac{9x^2}{2!} + \dfrac{81x^4}{4!} \ldots$

We neglect higher power terms like $x^4, x^6, \ldots$

$\displaystyle \lim_{x \to 0} \dfrac{1 - \left(1 - \frac{x^2}{2} \right)\left(1 - 2x^2\right)^{\frac{1}{2}}\left(1 - \frac{9x^2}{2}\right)^{\frac{1}{3}}}{x^2}$

Now as $x \to 0, (1+x)^n \approx (1+nx)$

$\displaystyle \lim_{x \to 0} \dfrac{1 - \left(1 - \frac{x^2}{2}\right)\left(1 - x^2\right)\left(1 - \frac{3x^2}{2}\right)}{x^2}$

$\Rightarrow \displaystyle \lim_{x \to 0} \dfrac{1 - \left(1 - \frac{x^2}{2} - x^2\right)\left(1 - \frac{3x^2}{2} \right)}{x^2}$

$\Rightarrow \displaystyle \lim_{x \to 0} \dfrac{1 - \left(1 - x^2 - \frac{x^2}{2} - \frac{3x^2}{2} \right)}{x^2}$

$\Rightarrow \displaystyle \lim_{x \to 0} \dfrac{(x^2)(\frac{1}{2} + 1 + \frac{3}{2})}{x^2}$

$\Rightarrow \displaystyle \dfrac{1}{2} + {1} + \dfrac{3}{2} = \boxed{3}$