Evaluate the following limits `lim_(x to 0)(sinax)/(sinbx), a,b ne 0`.

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} \) where \( a, b \neq 0 \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} \]

### Step 2: Use the standard limit
We know from the standard limit that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
We can manipulate our limit to use this result. We can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin(ax)}{ax} \cdot \frac{ax}{bx} \cdot \frac{bx}{\sin(bx)} \]

### Step 3: Split the limit
This gives us:
\[ \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \cdot \frac{a}{b} \cdot \frac{bx}{\sin(bx)} \right) \]
Now we can separate the limits:
\[ \lim_{x \to 0} \frac{\sin(ax)}{ax} \cdot \frac{a}{b} \cdot \lim_{x \to 0} \frac{bx}{\sin(bx)} \]

### Step 4: Evaluate each limit
Using the standard limit:
\[ \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{bx}{\sin(bx)} = 1 \]
Thus, we have:
\[ 1 \cdot \frac{a}{b} \cdot 1 = \frac{a}{b} \]

### Final Result
Therefore, the limit evaluates to:
\[ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \]