Evaluate the following limits :
`Lim_( xto 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 follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit expression:

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

### Step 2: Use the sine limit property
We know that as \( x \to 0 \), \( \frac{\sin kx}{kx} \to 1 \) for any constant \( k \). We can manipulate the expression to utilize this property.

We rewrite the numerator and denominator by multiplying and dividing by \( x \):

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

### Step 3: Factor out constants
Now we can factor out the constants from the limit:

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

### Step 4: Apply the limit property
Now we can apply the limit property \( \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \):

\[ = \frac{2 \cdot 1 + 6 \cdot 1}{5 \cdot 1 - 3 \cdot 1}. \]

### Step 5: Simplify the expression
This simplifies to:

\[ = \frac{2 + 6}{5 - 3} = \frac{8}{2} = 4. \]

### Final Answer
Thus, the limit is

\[ \lim_{x \to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} = 4. \]

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