`lim_{x\rightarrow 0}(cosx-cos3x)/(x(sin 3x-sinx))` is equal to

Options

A. (a) `-2/3`
B. (b) `1/3`
C. (c) `-1/2`
D. (d) 2

Answer

To solve the limit \( \lim_{x\rightarrow 0}\frac{\cos x - \cos 3x}{x(\sin 3x - \sin x)} \), we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \) approaches 0, both the numerator and denominator approach 0. This gives us the indeterminate form \( \frac{0}{0} \). Therefore, we need to simplify the expression.

**Hint:** When you encounter a \( \frac{0}{0} \) form, consider using trigonometric identities or L'Hôpital's Rule.

### Step 2: Use trigonometric identities
We can use the identities for the difference of cosines and sines:
- \( \cos a - \cos b = -2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \)
- \( \sin a - \sin b = 2 \cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \)

For the numerator \( \cos x - \cos 3x \):
- Let \( a = x \) and \( b = 3x \).
- Thus, \( \cos x - \cos 3x = -2 \sin\left(\frac{x + 3x}{2}\right)\sin\left(\frac{x - 3x}{2}\right) = -2 \sin(2x)\sin(-x) = 2 \sin(2x) \sin(x) \).

For the denominator \( \sin 3x - \sin x \):
- Let \( a = 3x \) and \( b = x \).
- Thus, \( \sin 3x - \sin x = 2 \cos\left(\frac{3x + x}{2}\right)\sin\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \sin(x) \).

### Step 3: Substitute back into the limit
Now we can rewrite the limit:
\[ \lim_{x\rightarrow 0} \frac{2 \sin(2x) \sin(x)}{x \cdot 2 \cos(2x) \sin(x)} \]
This simplifies to:
\[ \lim_{x\rightarrow 0} \frac{\sin(2x)}{x \cos(2x)} \]

**Hint:** When simplifying limits, try to cancel common factors.

### Step 4: Simplify further
We can separate the limit:
\[ \lim_{x\rightarrow 0} \frac{\sin(2x)}{2x} \cdot \frac{2}{\cos(2x)} \]
As \( x \) approaches 0, \( \frac{\sin(2x)}{2x} \) approaches 1 and \( \cos(2x) \) approaches 1.

### Step 5: Evaluate the limit
Thus, we have:
\[ \lim_{x\rightarrow 0} \frac{\sin(2x)}{2x} = 1 \quad \text{and} \quad \lim_{x\rightarrow 0} \frac{2}{\cos(2x)} = \frac{2}{1} = 2 \]
Therefore, the overall limit is:
\[ 1 \cdot 2 = 2 \]

### Conclusion
The limit evaluates to:
\[ \lim_{x\rightarrow 0}\frac{\cos x - \cos 3x}{x(\sin 3x - \sin x)} = 2 \]

The correct answer is option **d) 2**.

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