_ x → 0 ( ( 1 - ^ 2 ( 3x ) ^ 3 ( 4x ) )( ^ 3 ( 4x ) ( _ e (2x + 1))...

Given,

$\lim_{x \to 0}\left(\left(\dfrac{1 - \cos^{2} ( 3x )}{\cos^{3} ( 4x )}\right)\left(\dfrac{\sin^{3} ( 4x )}{{\left(\log_{e}\left(2x + 1\right)\right)}^{5}}\right)\right)$

Now we now that,

$\lim_{x \to 0}\dfrac{\sin x}{x} = 1, \lim_{x \to 0}\dfrac{1 - \cos x}{{x}^{2}} = \dfrac{1}{2} \text{ and } \lim_{x \to 0}\dfrac{\log \left(1 + x\right)}{x} = 1$

Now using the above formula we get,

$\lim_{x \to 0}\left(\left(\dfrac{1 - \cos^{2}3x}{\cos^{3}4x}\right)\left(\dfrac{\sin^{3}4x}{{\left(\log_{e}\left(2x + 1\right)\right)}^{5}}\right)\right)$

$= \lim_{x \to 0}\dfrac{\left(1 - \cos 3x\right)\left(1 + \cos 3x\right)9{x}^{2}}{\left(\cos^{3}4x\right)9{x}^{2}}\dfrac{{\left(\sin 4x\right)}^{3}{\left(64x\right)}^{3}{\left(2x\right)}^{5}}{\left(64{x}^{3}\right){\left(\log_{e} ( 2x + 1 )\right)}^{5}{\left(2x\right)}^{5}}$

$= \lim_{x \to 0}\dfrac{\left(\dfrac{1 - \cos 3x}{9{x}^{2}}\right)\left(1 + \cos 3x\right)}{\left(\cos^{3}4x\right)}\dfrac{{\left(\dfrac{\sin 4x}{4x}\right)}^{3}}{{\left(\dfrac{\log_{e} ( 2x + 1 )}{2x}\right)}^{5}} \times \dfrac{9 \times 64}{32}$

$= \dfrac{\left(\dfrac{1}{2}\right) \times \left(2\right)}{\left(1\right)}\dfrac{{\left(1\right)}^{3}}{{\left(1\right)}^{5}} \times \dfrac{9 \times 64}{32}$

$= \dfrac{9 \times 2}{1} = 18$