`lim _( x to 0) (tan 2x -x)/( 3x - sin x)=`

Answer

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

### Step 1: Identify the Form
When we substitute \( x = 0 \) directly into the limit, we get:
\[ \tan(2 \cdot 0) - 0 = 0 \quad \text{and} \quad 3 \cdot 0 - \sin(0) = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
We need to differentiate the numerator and the denominator.

### Step 3: Differentiate the Numerator and Denominator
1. **Numerator**: \( f(x) = \tan(2x) - x \)
- The derivative \( f'(x) = 2\sec^2(2x) - 1 \)

2. **Denominator**: \( g(x) = 3x - \sin x \)
- The derivative \( g'(x) = 3 - \cos x \)

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2\sec^2(2x) - 1}{3 - \cos x} \]

### Step 5: Substitute \( x = 0 \)
Now we substitute \( x = 0 \):
- For the numerator:
\[ 2\sec^2(2 \cdot 0) - 1 = 2\sec^2(0) - 1 = 2 \cdot 1 - 1 = 1 \]
- For the denominator:
\[ 3 - \cos(0) = 3 - 1 = 2 \]

### Step 6: Calculate the Limit
Now we can calculate the limit:
\[ \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\tan 2x - x}{3x - \sin x} = \frac{1}{2} \]