1. Rewrite the expression in a convenient form

Let $f(t)=t^3+2t^2+3\sin t.$ Then the given limit becomes $\lim_{x\to 0}\left(\frac{f(x+2\cos x)}{f(x+2)}\right)^{\frac{100}{x}}.$

We need a limit of the form $\left(1+\text{small quantity}\right)^{\frac{100}{x}},$ so we first study the ratio inside the bracket near $x=0$.

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2. Expand the two inputs of $f$ near $x=0$

Using $\cos x = 1-\frac{x^2}{2}+o(x^2),$ we get $x+2\cos x = x+2\left(1-\frac{x^2}{2}+o(x^2)\right)=2+x-x^2+o(x^2).$ Also, $x+2=2+x.$

Hence, $(x+2\cos x)-(x+2)=2\cos x-2=2(\cos x-1)=-x^2+o(x^2).$ So the two arguments of $f$ differ only by order $x^2$.

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3. Compare $f(x+2\cos x)$ and $f(x+2)$

By differentiability of $f$ near $2$, using linear approximation, $f(x+2\cos x)-f(x+2)=f'(2)\big[(x+2\cos x)-(x+2)\big]+o(x^2).$ Since the difference of arguments is $-x^2+o(x^2)$, we get $f(x+2\cos x)-f(x+2)=O(x^2).$

Also, $f(2)=2^3+2\cdot 2^2+3\sin 2=16+3\sin 2\neq 0.$ Therefore, $\frac{f(x+2\cos x)}{f(x+2)}=1+O(x^2).$

So the given limit is of the type $\left(1+O(x^2)\right)^{\frac{100}{x}}.$

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4. Take logarithm

Let $L=\lim_{x\to 0}\left(\frac{f(x+2\cos x)}{f(x+2)}\right)^{\frac{100}{x}}.$ Then $\ln L=\lim_{x\to 0}\frac{100}{x}\ln\left(1+O(x^2)\right).$ Using $\ln(1+u)\sim u \quad (u\to 0),$ with $u=O(x^2)$, we get $\ln\left(1+O(x^2)\right)=O(x^2).$ Hence, $\ln L=\lim_{x\to 0}\frac{100}{x}\cdot O(x^2)=\lim_{x\to 0} O(x)=0.$ Thus, $L=e^0=1.$

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5. Final answer

$\boxed{1}$

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6. Comparison with stored answer

Derived answer = $1$. Stored correct answer = $1$. They match.