Let `lim_(x to 0) ("sin" 2X)/(x) = a` and `lim_(x to 0) (3x)/(tan x) = b`, then a + b equals

Options

A. 5
B. 6
C. 0
D. 4

Answer

To solve the problem, we need to find the limits \( a \) and \( b \) and then add them together.

### Step-by-Step Solution

1. **Finding \( a \)**:
We need to evaluate the limit:
\[ a = \lim_{x \to 0} \frac{\sin(2x)}{x} \]
When we substitute \( x = 0 \), we get \( \frac{\sin(0)}{0} = \frac{0}{0} \), which is an indeterminate form. Therefore, we will apply L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
\[ a = \lim_{x \to 0} \frac{\frac{d}{dx}(\sin(2x))}{\frac{d}{dx}(x)} \]
The derivative of \( \sin(2x) \) is \( 2\cos(2x) \) and the derivative of \( x \) is \( 1 \). Thus, we have:
\[ a = \lim_{x \to 0} \frac{2\cos(2x)}{1} \]

3. **Evaluating the limit for \( a \)**:
Now, substituting \( x = 0 \):
\[ a = 2\cos(0) = 2 \cdot 1 = 2 \]

4. **Finding \( b \)**:
Next, we evaluate the limit:
\[ b = \lim_{x \to 0} \frac{3x}{\tan(x)} \]
Again, substituting \( x = 0 \) gives \( \frac{0}{0} \), which is also an indeterminate form. We will apply L'Hôpital's Rule here as well.

5. **Applying L'Hôpital's Rule for \( b \)**:
Differentiate the numerator and the denominator:
\[ b = \lim_{x \to 0} \frac{\frac{d}{dx}(3x)}{\frac{d}{dx}(\tan(x))} \]
The derivative of \( 3x \) is \( 3 \) and the derivative of \( \tan(x) \) is \( \sec^2(x) \). Thus, we have:
\[ b = \lim_{x \to 0} \frac{3}{\sec^2(x)} \]

6. **Evaluating the limit for \( b \)**:
Now, substituting \( x = 0 \):
\[ b = \frac{3}{\sec^2(0)} = \frac{3}{1} = 3 \]

7. **Finding \( a + b \)**:
Now we can find \( a + b \):
\[ a + b = 2 + 3 = 5 \]

### Final Answer
Thus, the value of \( a + b \) is \( \boxed{5} \).