Evaluate the following limits:
`lim_(xrarr0)((tan2x-x))/((3x-tanx))`

Explanation

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

### Step 1: Identify the limit form
First, we need to check the form of the limit as \( x \) approaches 0.

\[ \tan(2x) \to 0 \quad \text{and} \quad x \to 0 \quad \Rightarrow \quad \tan(2x) - x \to 0 - 0 = 0 \]

\[ 3x \to 0 \quad \text{and} \quad \tan(x) \to 0 \quad \Rightarrow \quad 3x - \tan(x) \to 0 - 0 = 0 \]

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(\tan(2x) - x) = 2\sec^2(2x) - 1 \]

**Denominator:**
\[ \frac{d}{dx}(3x - \tan(x)) = 3 - \sec^2(x) \]

### Step 3: Rewrite the limit
Now we can rewrite our limit using the derivatives:

\[ \lim_{x \to 0} \frac{2\sec^2(2x) - 1}{3 - \sec^2(x)} \]

### Step 4: Evaluate the limit
Now we substitute \( x = 0 \):

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

Substituting into the limit gives:

\[ \lim_{x \to 0} \frac{2(1) - 1}{3 - 1} = \frac{2 - 1}{3 - 1} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:

\[ \boxed{\frac{1}{2}} \]