Any other method

Method 1, (the long way): Using the identities

$\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$ and $\sin(2x) = 2\sin(x)\cos(x),$

we can rewrite the given function as

$f(x) = \dfrac{3\sin(x) - 4\sin^{3}(x) + 2A\sin(x)\cos(x) + B\sin(x)}{x^{5}} =$

$\dfrac{\sin(x)}{x} * \dfrac{3 - 4\sin^{2}(x) + 2A\cos(x) + B}{x^{4}}.$

Now since we want $\lim_{x \rightarrow 0} f(x)$ to exist, and since $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1,$ we will require that

$3 - 4\sin^{2}(x) + 2A\cos(x) + B = 0$ at $x = 0,$ i.e., that $3 + 2A + B = 0.$

This will then make $g(x) = \dfrac{3 - 4\sin^{2}(x) + 2A\cos(x) + B}{x^{4}}$

indeterminate at $x = 0.$ Applying L'Hopital's rule, we have that

$\lim_{x \rightarrow 0} \dfrac{3 - 4\sin^{2}(x) + 2A\cos(x) + B}{x^{4}} =$

$\lim_{x \rightarrow 0} \dfrac{-8\sin(x)\cos(x) - 2A\sin(x)}{4x^{3}} =$

$\lim_{x \rightarrow 0} \left( -\dfrac{1}{2}*\dfrac{\sin(x)}{x} * \dfrac{4\cos(x) + A}{x^{2}}\right),$ (i).

Again, since $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1,$ for this limit to exist we will require that $4\cos(x) + A = 0$ at $x = 0,$ i.e., that $A = -4.$

This in turn gives us that $B = -(3 + 2A) = 5.$

To make $f(x)$ continuous at $x = 0,$ we will need to then assign $f(0)$ the same value as

$\lim_{x \rightarrow 0} \dfrac{\sin(3x) - 4\sin(2x) + 5\sin(x)}{x^{5}}.$

To determine this limit, we just need to apply L'Hopital's rule to (i) once more, giving us that

$\lim_{x \rightarrow 0} f(x) = -\dfrac{1}{2} \lim_{x \rightarrow 0} \dfrac{4\cos(x) - 4}{x^{2}} =$

$-\dfrac{1}{2} \lim_{x \rightarrow 0} \dfrac{-4\sin(x)}{2x} = \lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1.$

Thus we can make $f(x)$ continuous at $x = 0$ if we assign the value $\boxed{f(0) = 1}.$

Method 2: Using the series expansion

$\sin(x) = x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} + O(x^{7}),$

we can rewrite $f(x)$ as

$f(x) = \dfrac{1}{x^{4}}(3 + 2A + B) - \dfrac{1}{6x^{2}}(27 + 8A + B) + \dfrac{1}{120}(243 + 32A + B) + O(x^{2}).$

For the limit to exist as $x \rightarrow 0$ we must have that

$3 + 2A + B = 0$ and that $27 + 8A + B = 0$

$\Longrightarrow 24 + 6A = 0 \Longrightarrow A = -4,$ and thus $B = -(3 + 2A) = 5.$

With these values, we would then have $\lim_{x \rightarrow 0} f(x) = \dfrac{1}{120}(243 + 32*(-4) + 5) = 1.$

Thus, again, to make $f(x)$ continuous at $x = 0$ we must assign $f(0) = \boxed{1}.$

My method may be same as that of Rishabh Deep Singh .

Let $g(x) = \sin(3x)+A\sin(2x)+B \sin (x)$ , for $f(0)$ to be continuous, $g(x)$ must have the form $g(x) = x^5 h(x)$ , so that $f(x) = \dfrac{g(x)}{x^5} = \dfrac{x^5h(x)}{x^5} = h(x)$ , which is continuous at $x=0$ .

$\begin{aligned} g(x) & = \sin(3x)+A\sin(2x)+B \sin (x) \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \small \left(3x - \frac{(3x)^3}{3!} + \frac{(3x)^5}{3!} -... \right) + A \left(2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{3!} -... \right) + B \left(x - \frac{x^3}{3!} + \frac{x^5}{3!} -...\right) \\ & = \small \color{#3D99F6}{3x - \frac{27x^3}{3!} + A \left(2x - \frac{8x^3}{3!} \right) + B \left(x - \frac{x^3}{3!} \right)} + x^5\left(\frac{3^5}{5!} - \frac{3^7x^2}{7!} + \frac{3^9x^4}{9!} -... \right) \\ & \quad \quad \small + Ax^5\left(\frac{2^5}{5!} - \frac{2^7x^2}{7!} + \frac{2^9x^4}{9!} -... \right) + Bx^5\left(\frac{1}{5!} - \frac{x^2}{7!} + \frac{x^4}{9!} -... \right) \end{aligned}$

To make $g(x) = x^5h(x)$ , then: $\small \color{#3D99F6}{3x - \frac{27x^3}{3!} + A \left(2x - \frac{8x^3}{3!} \right) + B \left(x - \frac{x^3}{3!} \right)} = 0$ and equating the coefficients, we have:

$\begin{cases} 3 + 2A + B = 0 \\ 27 + 8A + B = 0 \end{cases} \quad \Rightarrow A = -4, \space B = 5$

$\begin{aligned} \Rightarrow f(0) & = h(0) \\ & = \frac{3^5}{5!} + A \frac{2^5}{5!} + B \frac{1}{5!} \\ & = \frac{242 - 4\times 32 + 5\times 1}{5!} = \frac{120}{120} = \boxed{1} \end{aligned}$