Evaluate the following limits :
`lim_(x to 0) (sin 2x + sin 6x)/(sin 5x - sin 3x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x}, \]

we can use trigonometric identities to simplify the expression.

### Step 1: Apply the Sine Addition and Subtraction Identities

We will use the identities for sine addition and subtraction:

1. \(\sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)\)
2. \(\sin c - \sin d = 2 \cos\left(\frac{c+d}{2}\right) \sin\left(\frac{c-d}{2}\right)\)

For the numerator, let \(a = 6x\) and \(b = 2x\):

\[ \sin 2x + \sin 6x = 2 \sin\left(\frac{6x + 2x}{2}\right) \cos\left(\frac{6x - 2x}{2}\right) = 2 \sin(4x) \cos(2x). \]

For the denominator, let \(c = 5x\) and \(d = 3x\):

\[ \sin 5x - \sin 3x = 2 \cos\left(\frac{5x + 3x}{2}\right) \sin\left(\frac{5x - 3x}{2}\right) = 2 \cos(4x) \sin(x). \]

### Step 2: Substitute Back into the Limit

Now substituting these identities back into the limit gives us:

\[ \lim_{x \to 0} \frac{2 \sin(4x) \cos(2x)}{2 \cos(4x) \sin(x)}. \]

The factor of 2 cancels out:

\[ \lim_{x \to 0} \frac{\sin(4x) \cos(2x)}{\cos(4x) \sin(x)}. \]

### Step 3: Simplify the Limit

We can rewrite this limit as:

\[ \lim_{x \to 0} \left( \frac{\sin(4x)}{4x} \cdot \frac{4x}{\sin(x)} \cdot \cos(2x) \cdot \frac{1}{\cos(4x)} \right). \]

### Step 4: Evaluate Each Component

1. \(\lim_{x \to 0} \frac{\sin(4x)}{4x} = 1\) (using the standard limit).
2. \(\lim_{x \to 0} \frac{4x}{\sin(x)} = 4\) (using the standard limit).
3. \(\lim_{x \to 0} \cos(2x) = \cos(0) = 1\).
4. \(\lim_{x \to 0} \frac{1}{\cos(4x)} = \frac{1}{\cos(0)} = 1\).

### Step 5: Combine the Results

Putting it all together, we have:

\[ 1 \cdot 4 \cdot 1 \cdot 1 = 4. \]

Thus, the limit evaluates to:

\[ \boxed{4}. \]