Evaluate the following (10-17) limits :
`lim_(x to 0)(sin2x+3x)/(4x-sin5x)`

Answer

To evaluate the limit \(\lim_{x \to 0} \frac{\sin 2x + 3x}{4x - \sin 5x}\), we will follow these steps:

### Step 1: Substitute \(x = 0\)
First, we substitute \(x = 0\) into the limit:

\[ \sin(2 \cdot 0) + 3 \cdot 0 = \sin(0) + 0 = 0 \]
\[ 4 \cdot 0 - \sin(5 \cdot 0) = 0 - \sin(0) = 0 \]

Since both the numerator and denominator evaluate to 0, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

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

if the limit on the right exists. We will differentiate the numerator and the denominator with respect to \(x\).

#### Differentiate the Numerator:
\[ \text{Numerator: } \sin(2x) + 3x \]
Using the chain rule, the derivative is:
\[ \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \quad \text{and} \quad \frac{d}{dx}(3x) = 3 \]
Thus, the derivative of the numerator is:
\[ 2\cos(2x) + 3 \]

#### Differentiate the Denominator:
\[ \text{Denominator: } 4x - \sin(5x) \]
The derivative is:
\[ \frac{d}{dx}(4x) = 4 \quad \text{and} \quad \frac{d}{dx}(-\sin(5x)) = -5\cos(5x) \]
Thus, the derivative of the denominator is:
\[ 4 - 5\cos(5x) \]

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

\[ \lim_{x \to 0} \frac{2\cos(2x) + 3}{4 - 5\cos(5x)} \]

### Step 4: Substitute \(x = 0\) Again
Now we substitute \(x = 0\) into the new limit:

\[ \text{Numerator: } 2\cos(2 \cdot 0) + 3 = 2\cos(0) + 3 = 2 \cdot 1 + 3 = 5 \]
\[ \text{Denominator: } 4 - 5\cos(5 \cdot 0) = 4 - 5\cos(0) = 4 - 5 \cdot 1 = 4 - 5 = -1 \]

### Step 5: Final Calculation
Now we can compute the limit:

\[ \lim_{x \to 0} \frac{5}{-1} = -5 \]

Thus, the final answer is:

\[ \boxed{-5} \]