`lim_(xrarr0) (sin 3 x)/(2x)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin 3x}{2x} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{\sin 3x}{2x} \]

### Step 2: Multiply by a factor to simplify
To simplify the expression, we can multiply the numerator and the denominator by 3:
\[ \lim_{x \to 0} \frac{\sin 3x}{2x} = \lim_{x \to 0} \frac{3 \sin 3x}{6x} \]

### Step 3: Use the standard limit
Now, we can separate the limit:
\[ \lim_{x \to 0} \frac{3 \sin 3x}{6x} = \frac{3}{6} \lim_{x \to 0} \frac{\sin 3x}{3x} \]
This simplifies to:
\[ \frac{1}{2} \lim_{x \to 0} \frac{\sin 3x}{3x} \]

### Step 4: Apply the standard limit result
We know from calculus that:
\[ \lim_{u \to 0} \frac{\sin u}{u} = 1 \]
In our case, as \( x \to 0 \), \( 3x \to 0 \). Thus:
\[ \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \]

### Step 5: Substitute back into the limit
Now substituting this result back into our expression gives:
\[ \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin 3x}{2x} = \frac{3}{2} \]

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