_ x → 0 ( (x + 2 x) ^ 3 + 2 (x + 2 x) ^ 2 + 3 (x + 2 x) (x + 2) ^ 3...

Given,$\lim_{x \to 0}\left(\dfrac{{\left(x + 2\cos x\right)}^{3} + 2{\left(x + 2\cos x\right)}^{2} + 3\sin \left(x + 2\cos x\right)}{{\left(x + 2\right)}^{3} + 2{\left(x + 2\right)}^{2} + 3\sin \left(x + 2\right)}\right)\dfrac{100}{x}$

Now by putting the value of limit we can see it is in form ${1}^{\infty}$

Now we know that,

$\lim_{x \to 0}{\left(f\left(x\right)\right)}^{g\left(x\right)}$ when is in form of ${1}^{\infty}$, we can write this as ${e}^{\lim_{x \to 0}\left(f\left(x\right) - 1\right)g\left(x\right)}$

Now using the above rule we get,

$= {e}^{\lim_{x \to 0}\left[\left(\dfrac{{\left(x + 2\cos x\right)}^{3} + 2{\left(x + 2\cos x\right)}^{2} + 3\sin \left(x + 2\cos x\right)}{{\left(x + 2\right)}^{3} + 2{\left(x + 2\right)}^{2} + 3\sin \left(x + 2\right)}\right) - 1\right] \times \dfrac{100}{x}}$

$= {e}^{\lim_{x \to 0}\left[\dfrac{100}{x}\left[\dfrac{{\left(x + 2\cos x\right)}^{3} + 2{\left(x + 2\cos x\right)}^{2} + 3\sin \left(x + 2\cos x\right) - \left({\left(x + 2\right)}^{3} + 2{\left(x + 2\right)}^{2} + 3\sin \left(x + 2\right)\right)}{{\left(x + 2\right)}^{3} + 2{\left(x + 2\right)}^{2} + 3\sin \left(x + 2\right)}\right]\right]}$

$= {e}^{\lim_{x \to 0}\dfrac{100}{x}\left[\left(\dfrac{{\left(x + 2\cos x\right)}^{3} - {\left(x + 2\right)}^{3} + 2{\left(x + 2\cos x\right)}^{2} - 2{\left(x + 2\right)}^{2} + 3\sin \left(x + 2\cos x\right) - 3\sin \left(x + 2\right)}{8 + 8 + 3\sin 2}\right)\right]}$

Using L-hospital method we get,

$= {e}^{\dfrac{100}{16 + 3\sin 2}\lim_{x \to 0}\dfrac{3{\left(x + 2\cos x\right)}^{2} \times \left(1 + 2\sin x\right) - 3{\left(x + 2\right)}^{2} + 4\left(x + 2\cos x\right) \times \left(1 - 2\sin x\right) - 4\left(x + 2\right) + 3\cos \left(x + 2\cos x\right) \times \left(1 - 2\sin x\right) - 3\cos \left(x + 2\right)}{1}}$

$= {e}^{\dfrac{100}{16 + 3\sin 2}\left(\dfrac{12 - 3\left(4\right) + 8 \times 1 - 8 + 3\cos 2 - 3\cos 2}{1}\right)}$

$= {e}^{\dfrac{100}{16 + 3\sin 2} \times 0}$

$= {e}^{0} = 1$