Evaluate the following limits:
`lim_(xrarr0)((sin3x+sin5x))/((sin6x-sin4x))`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin(3x) + \sin(5x)}{\sin(6x) - \sin(4x)}, \]

we can follow these steps:

### Step 1: Rewrite the limit using the property of sine
We know that \(\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1\) for any constant \(k\). We can use this property to rewrite the sine functions in the limit.

### Step 2: Multiply and divide by appropriate factors
We can multiply the numerator and the denominator by \(3x\), \(5x\), \(6x\), and \(4x\) respectively to facilitate the limit evaluation:

\[ \lim_{x \to 0} \frac{\sin(3x) + \sin(5x)}{\sin(6x) - \sin(4x)} = \lim_{x \to 0} \frac{\left(\sin(3x) \cdot \frac{3x}{3x} + \sin(5x) \cdot \frac{5x}{5x}\right)}{\left(\sin(6x) \cdot \frac{6x}{6x} - \sin(4x) \cdot \frac{4x}{4x}\right)}. \]

This gives us:

\[ = \lim_{x \to 0} \frac{\left(\frac{\sin(3x)}{3x} \cdot 3x + \frac{\sin(5x)}{5x} \cdot 5x\right)}{\left(\frac{\sin(6x)}{6x} \cdot 6x - \frac{\sin(4x)}{4x} \cdot 4x\right)}. \]

### Step 3: Factor out \(x\)
Now we can factor out \(x\) from both the numerator and the denominator:

\[ = \lim_{x \to 0} \frac{x\left(\frac{\sin(3x)}{3x} \cdot 3 + \frac{\sin(5x)}{5x} \cdot 5\right)}{x\left(\frac{\sin(6x)}{6x} \cdot 6 - \frac{\sin(4x)}{4x} \cdot 4\right)}. \]

### Step 4: Cancel \(x\)
We can cancel \(x\) from the numerator and denominator:

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

### Step 5: Apply the limit
Now we can apply the limit as \(x\) approaches \(0\):

\[ = \frac{3 \cdot 1 + 5 \cdot 1}{6 \cdot 1 - 4 \cdot 1} = \frac{3 + 5}{6 - 4} = \frac{8}{2} = 4. \]

### Final Answer
Thus, the limit evaluates to:

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