`Lt_(x rarr0)(tan3x)/(2x)` is equal to

Options

A. `(2)/(3)`
B. `(3)/(2)`
C. `(9)/(4)`
D. none of these

Answer

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

### Step 1: Direct Substitution
First, we will try to directly substitute \( x = 0 \) into the limit.

\[ \frac{\tan(3 \cdot 0)}{2 \cdot 0} = \frac{\tan(0)}{0} = \frac{0}{0} \]

This is an indeterminate form \( \frac{0}{0} \), so we cannot evaluate the limit directly.

**Hint:** If you encounter \( \frac{0}{0} \) when substituting, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we will apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

Here, \( f(x) = \tan(3x) \) and \( g(x) = 2x \).

### Step 3: Differentiate the Numerator and Denominator
Now, we differentiate the numerator and the denominator:

- The derivative of \( f(x) = \tan(3x) \) is:
\[ f'(x) = 3 \sec^2(3x) \]
(using the chain rule, where the derivative of \( \tan(u) \) is \( \sec^2(u) \) and \( u = 3x \) gives us an additional factor of 3).

- The derivative of \( g(x) = 2x \) is:
\[ g'(x) = 2 \]

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{\tan(3x)}{2x} = \lim_{x \to 0} \frac{3 \sec^2(3x)}{2} \]

### Step 5: Substitute \( x = 0 \) Again
Now we substitute \( x = 0 \):

\[ = \frac{3 \sec^2(3 \cdot 0)}{2} = \frac{3 \sec^2(0)}{2} \]

Since \( \sec(0) = 1 \), we have:

\[ \sec^2(0) = 1 \]

Thus, the limit becomes:

\[ = \frac{3 \cdot 1}{2} = \frac{3}{2} \]

### Final Answer
Therefore, the limit is:

\[ \lim_{x \to 0} \frac{\tan(3x)}{2x} = \frac{3}{2} \]

---