Question:

Step 1: Determine the function $f(x)$. The given integral is $$ \int \frac{3x\sin^2 x\cos x - 3\sin^3 x}{x^4}\,dx = \int \frac{3\sin^2 x(x\cos x - \sin x)}{x^4}\,dx $$ This integrand is the derivative of $\frac{\sin^3 x}{x^3}$ with respect to $x$. $$ \frac{d}{dx}\left(\frac{\sin^3 x}{x^3}\right) = \frac{(3\sin^2 x \cos x)x^3 - (\sin^3 x)(3x^2)}{(x^3)^2} = \frac{3x^3\sin^2 x \cos x - 3x^2\sin^3 x}{x^6} = \frac{3\sin^2 x(x\cos x - \sin x)}{x^4} $$ Thus, $f(x) = \frac{\sin^3 x}{x^3}$. Step 2: Verify the limit condition for $f(x)$. The condition $\displaystyle\lim_{x \to 0} f(x) = 1$ is checked: $$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\sin^3 x}{x^3} = \lim_{x\to 0} \left(\frac{\sin x}{x}\right)^3 = 1^3 = 1 $$ The condition is satisfied. Step 3: Evaluate the first limit expression. Consider the limit $\displaystyle\lim_{x\to 0}\frac{\int_0^x tf(t)\,dt - 2x^2}{1-\cos x}$. This is an indeterminate form $\frac{0}{0}$. Applying L'Hôpital's Rule: $$ \lim_{x\to 0}\frac{\frac{d}{dx}\left(\int_0^x tf(t)\,dt - 2x^2\right)}{\frac{d}{dx}(1-\cos x)} = \lim_{x\to 0}\frac{xf(x) - 4x}{\sin x} $$ Substitute $f(x) = \frac{\sin^3 x}{x^3}$: $$ = \lim_{x\to 0}\frac{x\left(\frac{\sin^3 x}{x^3}\right) - 4x}{\sin x} = \lim_{x\to 0}\frac{\frac{\sin^3 x}{x^2} - 4x}{\sin x} $$ Divide the numerator and denominator by $x$: $$ = \lim_{x\to 0}\frac{\frac{\sin^3 x}{x^3} - 4}{\frac{\sin x}{x}} = \frac{1^3 - 4}{1} = \frac{1 - 4}{1} = -3 $$ Step 4: Evaluate the second limit expression. Consider the limit $\displaystyle\lim_{x\to 0}\frac{(f(x))^{1/3} - 1}{x^2}$. First, calculate $(f(x))^{1/3}$: $$ (f(x))^{1/3} = \left(\frac{\sin^3 x}{x^3}\right)^{1/3} = \frac{\sin x}{x} $$ Substitute this into the limit expression: $$ \lim_{x\to 0}\frac{\frac{\sin x}{x} - 1}{x^2} $$ Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + O(x^5)$, we have $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4)$. $$ \lim_{x\to 0}\frac{\left(1 - \frac{x^2}{6} + O(x^4)\right) - 1}{x^2} = \lim_{x\to 0}\frac{-\frac{x^2}{6} + O(x^4)}{x^2} = \lim_{x\to 0}\left(-\frac{1}{6} + O(x^2)\right) = -\frac{1}{6} $$ Step 5: Evaluate the definite integral involving $h(x)$. Let $h(x) = x \cdot \sqrt[3]{f(x)}$. Substitute $f(x) = \frac{\sin^3 x}{x^3}$: $$ h(x) = x \cdot \left(\frac{\sin^3 x}{x^3}\right)^{1/3} = x \cdot \frac{\sin x}{x} = \sin x $$ Now, evaluate the integral $\displaystyle\int_0^{\pi} h^4(x)\,dx$: $$ \int_0^{\pi} \sin^4 x\,dx $$ Use the power-reducing identity $\sin^2 x = \frac{1-\cos 2x}{2}$: $$ \int_0^{\pi} \left(\frac{1-\cos 2x}{2}\right)^2 dx = \int_0^{\pi} \frac{1 - 2\cos 2x + \cos^2 2x}{4}\,dx $$ Use another power-reducing identity $\cos^2 2x = \frac{1+\cos 4x}{2}$: $$ = \frac{1}{4}\int_0^{\pi} \left(1 - 2\cos 2x + \frac{1+\cos 4x}{2}\right)\,dx $$ $$ = \frac{1}{4}\int_0^{\pi} \left(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x\right)\,dx $$ $$ = \frac{1}{4}\left[\frac{3}{2}x - \sin 2x + \frac{1}{8}\sin 4x\right]_0^{\pi} $$ $$ = \frac{1}{4}\left[\left(\frac{3}{2}\pi - \sin(2\pi) + \frac{1}{8}\sin(4\pi)\right) - \left(0 - \sin(0) + \frac{1}{8}\sin(0)\right)\right] $$ $$ = \frac{1}{4}\left[\frac{3}{2}\pi - 0 + 0 - 0\right] = \frac{3\pi}{8} $$