Evaluate the following (10-17) limits :
`lim_(x to 0)(tan3x+x)/(2x+sin4x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\tan(3x) + x}{2x + \sin(4x)}, \]

we start by substituting \(x = 0\):

1. **Substitution**:
\[ \tan(3 \cdot 0) + 0 = \tan(0) + 0 = 0 + 0 = 0, \]
and
\[ 2 \cdot 0 + \sin(4 \cdot 0) = 0 + \sin(0) = 0 + 0 = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

- Differentiate the numerator:
\[ \frac{d}{dx}(\tan(3x) + x) = 3\sec^2(3x) + 1. \]

- Differentiate the denominator:
\[ \frac{d}{dx}(2x + \sin(4x)) = 2 + 4\cos(4x). \]

3. **Re-evaluating the limit**:
Now we can rewrite our limit using the derivatives:
\[ \lim_{x \to 0} \frac{3\sec^2(3x) + 1}{2 + 4\cos(4x)}. \]

4. **Substituting \(x = 0\) again**:
- For the numerator:
\[ 3\sec^2(3 \cdot 0) + 1 = 3\sec^2(0) + 1 = 3 \cdot 1 + 1 = 3 + 1 = 4. \]

- For the denominator:
\[ 2 + 4\cos(4 \cdot 0) = 2 + 4\cos(0) = 2 + 4 \cdot 1 = 2 + 4 = 6. \]

5. **Final calculation**:
Now we can compute the limit:
\[ \lim_{x \to 0} \frac{4}{6} = \frac{2}{3}. \]

Thus, the final answer is

\[ \frac{2}{3}. \]