We want to calculate the limit:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \]
In this proof, we will explore two main methods to calculate the limit in question. The first approach is based on using the well-known limit property that states \(\lim_{x \to 0} \displaystyle \frac{\sin(x)}{x} = 1\), applied appropriately. The second approach uses Taylor series to approximate \(\sin(x)\) in a neighborhood of \(x = 0\), allowing us to obtain the result analytically. Let's now examine both demonstrations in detail.
Approach using the limit property
We write the limit as:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \cdot \frac{bx}{\sin(bx)} \cdot \frac{a}{b} \right) \]
We apply the property \(\lim_{x \to 0} \displaystyle \frac{\sin(x)}{x} = 1\) to \(\sin(ax)\) and \(\sin(bx)\):
\[ \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{bx}{\sin(bx)} = 1 \]
The limit then simplifies to:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \]
Approach using Taylor series
The Taylor series for \(\sin(x)\) around \(x = 0\) is:
\[ \sin(x) = x - \frac{x^3}{6} + O(x^5) \]
Using this expansion, we can write \(\sin(ax)\) and \(\sin(bx)\) as:
\[ \sin(ax) = ax - \frac{(ax)^3}{6} + O(x^5) \quad \text{and} \quad \sin(bx) = bx - \frac{(bx)^3}{6} + O(x^5) \]
Now, consider the ratio \(\displaystyle \frac{\sin(ax)}{\sin(bx)}\):
\[ \frac{\sin(ax)}{\sin(bx)} = \frac{ax - \displaystyle \frac{(ax)^3}{6} + O(x^5)}{bx - \displaystyle \frac{(bx)^3}{6} + O(x^5)} \]
As \(x \to 0\), the terms involving \(x^2\) and higher powers tend to zero. Therefore, the limit becomes:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \]