Evaluate `lim_(x to 0) ("sin"^(2) ax)/("sin"^(2) bx)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin^2(ax)}{\sin^2(bx)}, \]

we can follow these steps:

### Step 1: Rewrite the limit
We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin^2(ax)}{\sin^2(bx)} = \lim_{x \to 0} \left( \frac{\sin(ax)}{\sin(bx)} \right)^2. \]

### Step 2: Use the limit property
We know that as \( x \to 0 \), \( \frac{\sin(kx)}{kx} \to 1 \) for any constant \( k \). Therefore, we can express \( \sin(ax) \) and \( \sin(bx) \) in terms of their arguments:

\[ \frac{\sin(ax)}{ax} \cdot \frac{ax}{bx} \cdot \frac{bx}{\sin(bx)}. \]

### Step 3: Substitute and simplify
Substituting this into our limit, we have:

\[ \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \cdot \frac{ax}{bx} \cdot \frac{bx}{\sin(bx)} \right)^2. \]

### Step 4: Evaluate each part of the limit
As \( x \to 0 \):

- \( \frac{\sin(ax)}{ax} \to 1 \)
- \( \frac{bx}{\sin(bx)} \to 1 \)

Thus, we can simplify our limit to:

\[ \lim_{x \to 0} \left( 1 \cdot \frac{a}{b} \cdot 1 \right)^2 = \left( \frac{a}{b} \right)^2. \]

### Step 5: Final result
Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\sin^2(ax)}{\sin^2(bx)} = \left( \frac{a}{b} \right)^2. \]