Number of Solutions of Trigonometric Limit Equation

To evaluate the given function $f(x)$, we consider the limit: $f(x) = \lim_{y \rightarrow 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^3}$

For a fixed non-zero $x \in \mathbf{R}$, let $u = xy$. As $y \to 0$, $u \to 0$. We can rewrite the limit by multiplying and dividing by appropriate powers of $u$: $f(x) = \lim_{y \rightarrow 0} \left( \frac{1 - \cos(xy)}{(xy)^2} \right) \cdot \left( \frac{\tan(xy)}{xy} \right) \cdot \left( \frac{(xy)^3}{y^3} \right)$

Using the standard limits: $\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac{1}{2} \quad \text{and} \quad \lim_{u \to 0} \frac{\tan u}{u} = 1$

We obtain: $f(x) = \frac{1}{2} \cdot 1 \cdot x^3 = \frac{x^3}{2}$

For $x = 0$, $f(0) = \lim_{y \rightarrow 0} \frac{0}{y^3} = 0$, which also satisfies $f(0) = \frac{0^3}{2} = 0$. Thus, $f(x) = \frac{x^3}{2}$ for all $x \in \mathbf{R}$.

Next, we need to find the number of solutions to the equation $f(x) = \sin x$: $\frac{x^3}{2} = \sin x \implies x^3 - 2\sin x = 0$

Let $g(x) = x^3 - 2\sin x$. We seek the number of real roots of $g(x) = 0$.

1. Symmetry: Since $g(-x) = (-x)^3 - 2\sin(-x) = -x^3 + 2\sin x = -g(x)$, $g(x)$ is an odd function. Thus, $x = 0$ is a solution since $g(0) = 0^3 - 2\sin(0) = 0$. Any non-zero real solutions must occur in pairs of opposite signs.

2. Positive Real Solutions ($x > 0$):

   • For $x > \sqrt[3]{2} \approx 1.26$: $x^3 > 2 \ge 2\sin x \implies g(x) = x^3 - 2\sin x > 0$ Therefore, there are no solutions for $x > \sqrt[3]{2}$.

   • For $x \in (0, \sqrt[3]{2}]$: $g'(x) = 3x^2 - 2\cos x$ $g''(x) = 6x + 2\sin x$ Since $x \in (0, \sqrt[3]{2}] \subset (0, \pi)$, both $6x > 0$ and $\sin x > 0$, which means $g''(x) > 0$. Hence, $g'(x)$ is strictly increasing on $[0, \sqrt[3]{2}]$.

     Evaluating $g'(x)$ at the endpoints: $g'(0) = -2 < 0$ $g'(1) = 3 - 2\cos(1) \approx 3 - 2(0.54) = 1.92 > 0$

     Since $g'(x)$ is strictly increasing and changes sign, $g'(x) = 0$ has a unique root $\alpha \in (0, 1)$.

     • For $x \in (0, \alpha)$, $g'(x) < 0$, so $g(x)$ strictly decreases from $g(0) = 0$ to a negative minimum $g(\alpha) < 0$.
     • For $x > \alpha$, $g'(x) > 0$, so $g(x)$ strictly increases.

     At $x = \sqrt[3]{2}$: $g(\sqrt[3]{2}) = 2 - 2\sin(\sqrt[3]{2}) > 0 \quad (\text{since } \sin(1.26 \text{ rad}) < 1)$

     By the Intermediate Value Theorem, $g(x)$ increases from $g(\alpha) < 0$ to $g(\sqrt[3]{2}) > 0$, crossing zero exactly once in the interval $(\alpha, \sqrt[3]{2}]$. Thus, there is exactly 1 positive real solution.

3. Negative Real Solutions ($x < 0$): By the odd symmetry of $g(x)$, there is also exactly 1 negative real solution.

Combining all cases, the total number of real solutions is: $1 \text{ (negative)} + 1 \text{ (zero)} + 1 \text{ (positive)} = 3$

Thus, the correct option is C.