`lim_(x to 0) (sin 3x)/(x)` is equal to

Options

A. 1
B. 3
C. 2
D. 0

Answer

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

### Step 1: Recognize the standard limit
We know from trigonometric limits that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
This is a crucial formula that we will use in our solution.

### Step 2: Rewrite the limit
In our case, we have \( \sin(3x) \) in the numerator. To use the standard limit, we need to express the limit in a form that resembles \( \frac{\sin(u)}{u} \). We can do this by manipulating the expression:
\[ \lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 \]
Here, we multiplied and divided by 3 to balance the equation.

### Step 3: Apply the limit
Now we can apply the limit:
\[ \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 \]
As \( x \to 0 \), \( 3x \to 0 \) as well. Therefore, we can use the standard limit:
\[ \lim_{x \to 0} \frac{\sin(3x)}{3x} = 1 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{\sin(3x)}{x} = 1 \cdot 3 = 3 \]

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

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