`lim_(h -> 0) (sin(a+3h)-3sin(a+2h)+3sin(a+h)-sina)/h^3 =`

Options

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

Answer

To solve the limit

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

we will follow these steps:

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

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

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

This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** If substituting directly gives \( \frac{0}{0} \), consider applying L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule. We differentiate the numerator and the denominator separately.

**Numerator:**
- The derivative of \( \sin(a + 3h) \) is \( 3\cos(a + 3h) \).
- The derivative of \( -3\sin(a + 2h) \) is \( -6\cos(a + 2h) \).
- The derivative of \( 3\sin(a + h) \) is \( 3\cos(a + h) \).
- The derivative of \( -\sin(a) \) is \( 0 \).

Thus, the derivative of the numerator becomes:

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

**Denominator:**
The derivative of \( h^3 \) is \( 3h^2 \).

Now we rewrite the limit:

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

**Hint:** After applying L'Hôpital's Rule, check if you still have an indeterminate form.

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

Substituting \( h = 0 \) again gives:

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

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

**Hint:** If you still get \( \frac{0}{0} \), you need to differentiate again.

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

Differentiate the numerator and denominator again.

**Numerator:**
- The derivative of \( 3\cos(a + 3h) \) is \( -9\sin(a + 3h) \).
- The derivative of \( -6\cos(a + 2h) \) is \( 12\sin(a + 2h) \).
- The derivative of \( 3\cos(a + h) \) is \( -3\sin(a + h) \).

So the new numerator is:

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

**Denominator:**
The derivative of \( 3h^2 \) is \( 6h \).

Now we rewrite the limit:

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

**Hint:** Substitute \( h = 0 \) again to check if you still have an indeterminate form.

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

Substituting \( h = 0 \) gives:

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

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

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

Differentiate the numerator and denominator once more.

**Numerator:**
- The derivative of \( -9\sin(a + 3h) \) is \( -27\cos(a + 3h) \).
- The derivative of \( 12\sin(a + 2h) \) is \( 24\cos(a + 2h) \).
- The derivative of \( -3\sin(a + h) \) is \( -3\cos(a + h) \).

So the new numerator is:

\[ -27\cos(a + 3h) + 24\cos(a + 2h) - 3\cos(a + h). \]

**Denominator:**
The derivative of \( 6h \) is \( 6 \).

Now we rewrite the limit:

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

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

Substituting \( h = 0 \) gives:

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

So the limit evaluates to:

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

### Final Answer

Thus, the limit is:

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