The value of \( \lim_{x\rightarrow\infty}\frac{x^{3}+2x^{2}\sin x-4x \cos x}{\sqrt{(3x^{2}+2x \cos x)^{3}}} = \)
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For infinity-type limits, bounded oscillating terms like \( \sin x \) and \( \cos x \) can be treated as constants because they are completely dominated by high powers of \( x \) as \( x \) grows infinitely large.
Step 1: Spot the leading powers. As $x\to\infty$, the term that grows fastest decides the limit. In the numerator the top power is $x^3$; inside the denominator's bracket the top power is $x^2$. Step 2: Factor $x^3$ from the numerator. \[ x^3+2x^2\sin x-4x\cos x=x^3\left(1+\frac{2\sin x}{x}-\frac{4\cos x}{x^2}\right) \] Step 3: Simplify the denominator. Factor $x^2$ inside the bracket: \[ (3x^2+2x\cos x)^3=x^6\left(3+\frac{2\cos x}{x}\right)^3 \] Step 4: Take the square root. \[ \sqrt{x^6\left(3+\tfrac{2\cos x}{x}\right)^3}=x^3\left(3+\frac{2\cos x}{x}\right)^{3/2} \] Step 5: Use the vanishing terms. Since $\frac{\sin x}{x}\to0$ and $\frac{\cos x}{x}\to0$ as $x\to\infty$, the brackets approach their constant parts. Step 6: Evaluate the ratio. \[ \lim_{x\to\infty}\frac{x^3(1+0-0)}{x^3(3+0)^{3/2}}=\frac{1}{3^{3/2}}=\frac{1}{3\sqrt{3}} \] \[ \boxed{\tfrac{1}{3\sqrt{3}}} \]