Question
Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^{3}}$. Then the number of solutions of the equation $f(x) = \sin x, \ x \in \mathbf{R}$ is :
The correct answer is
3
To solve for the number of solutions to the equation \( f(x) = \sin x \) where \( f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^{3}} \) and \( x \in \mathbf{R} \), we need to evaluate the function \( f(x) \) first.
- Begin by analyzing the expression \( \lim_{y \to 0} \frac{(1 - \cos(xy)) \tan(xy)}{y^{3}} \).
- Use the trigonometric approximation for small angles: \(1 - \cos(xy) \approx \frac{(xy)^2}{2}\) and \(\tan(xy) \approx xy\).
- Substitute these approximations into the limit: \(\lim_{y \to 0} \frac{\left(\frac{(xy)^2}{2}\right)(xy)}{y^3} = \lim_{y \to 0} \frac{x^3y^3}{2y^3} = \frac{x^3}{2}\).
- Thus, we have \(f(x) = \frac{x^3}{2}\).
- Now, solve for \(\frac{x^3}{2} = \sin x\):
- Rearrange to \(x^3 = 2 \sin x\).
- Find the number of solutions for this equation in real numbers. Consider the graphs of \(x^3\) and \(2\sin x\):
- The function \(x^3\) is a cubic curve, and \(2 \sin x\) is a sinusoidal wave with amplitude 2.
- Since \(2 \sin x\) is bounded between -2 and 2, and \(x^3)\)is unbounded, focus on the intersection within this range.
- Graphically, visualize or draw the graphs of both functions to identify the intersection points. The crossover points indicate the solutions to the equation.
- Upon observing or utilizing calculus or numerical methods to check for points of intersection, there are three distinct real solutions for the equation.
Therefore, the number of solutions of the equation \( f(x) = \sin x \) is: 3.
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