Evaluate
`lim_(xrarr 0) (sin(3x+a)-3 sin(2x+a)+3sin(x+a)-sina)/(x^(3))`

Options

A. 0
B. cos a
C. `- cos a `
D. sin a

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin(3x + a) - 3\sin(2x + a) + 3\sin(x + a) - \sin a}{x^3}, \]

we will follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the limit expression:

\[ \sin(3(0) + a) - 3\sin(2(0) + a) + 3\sin(0 + a) - \sin a = \sin a - 3\sin a + 3\sin a - \sin a = 0. \]

So, we have:

\[ \frac{0}{0} \]

This indicates that we have an indeterminate form, and we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have \( \frac{0}{0} \), we differentiate the numerator and the denominator.

**Numerator:**
\[ \frac{d}{dx} \left( \sin(3x + a) - 3\sin(2x + a) + 3\sin(x + a) - \sin a \right) = 3\cos(3x + a) - 6\cos(2x + a) + 3\cos(x + a). \]

**Denominator:**
\[ \frac{d}{dx}(x^3) = 3x^2. \]

Now, we rewrite the limit:

\[ \lim_{x \to 0} \frac{3\cos(3x + a) - 6\cos(2x + a) + 3\cos(x + a)}{3x^2}. \]

### Step 3: Substitute \( x = 0 \) Again

Substituting \( x = 0 \) again gives:

\[ 3\cos(3(0) + a) - 6\cos(2(0) + a) + 3\cos(0 + a) = 3\cos a - 6\cos a + 3\cos a = 0. \]

We again have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator again.

**Numerator:**
\[ \frac{d}{dx} \left( 3\cos(3x + a) - 6\cos(2x + a) + 3\cos(x + a) \right) = -9\sin(3x + a) + 12\sin(2x + a) - 3\sin(x + a). \]

**Denominator:**
\[ \frac{d}{dx}(3x^2) = 6x. \]

Now, we rewrite the limit:

\[ \lim_{x \to 0} \frac{-9\sin(3x + a) + 12\sin(2x + a) - 3\sin(x + a)}{6x}. \]

### Step 5: Substitute \( x = 0 \) Again

Substituting \( x = 0 \) gives:

\[ -9\sin(3(0) + a) + 12\sin(2(0) + a) - 3\sin(0 + a) = -9\sin a + 12\sin a - 3\sin a = 0. \]

We again have \( \frac{0}{0} \), so we apply L'Hôpital's Rule one more time.

### Step 6: Apply L'Hôpital's Rule One More Time

Differentiate the numerator and denominator again.

**Numerator:**
\[ \frac{d}{dx} \left( -9\sin(3x + a) + 12\sin(2x + a) - 3\sin(x + a) \right) = -27\cos(3x + a) + 24\cos(2x + a) - 3\cos(x + a). \]

**Denominator:**
\[ \frac{d}{dx}(6x) = 6. \]

Now, we rewrite the limit:

\[ \lim_{x \to 0} \frac{-27\cos(3x + a) + 24\cos(2x + a) - 3\cos(x + a)}{6}. \]

### Step 7: Substitute \( x = 0 \) One Last Time

Substituting \( x = 0 \) gives:

\[ -27\cos(0 + a) + 24\cos(0 + a) - 3\cos(0 + a) = (-27 + 24 - 3)\cos a = -6\cos a. \]

Thus, the limit becomes:

\[ \frac{-6\cos a}{6} = -\cos a. \]

### Final Answer

The final result is:

\[ \lim_{x \to 0} \frac{\sin(3x + a) - 3\sin(2x + a) + 3\sin(x + a) - \sin a}{x^3} = -\cos a. \]