If `lim_(x rarr 0) (cos4x+a cos2x+b)/x^4` is finite then the value of `a,b` respectively

Options

A. `(5,-4)`
B. `(-5,-4)`
C. `(-4,3)`
D. `(4,5)`

Explanation

To solve the limit problem, we need to find the values of \( a \) and \( b \) such that the limit

\[ \lim_{x \to 0} \frac{\cos(4x) + a \cos(2x) + b}{x^4} \]

is finite. Let's go through the steps to find \( a \) and \( b \).

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

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

\[ \cos(4 \cdot 0) + a \cos(2 \cdot 0) + b = 1 + a + b \]

The denominator becomes:

\[ 0^4 = 0 \]

Thus, we have:

\[ \frac{1 + a + b}{0} \]

For the limit to be finite, the numerator must also equal zero when \( x \) approaches 0. Therefore, we set:

\[ 1 + a + b = 0 \]

### Step 2: Differentiate using L'Hôpital's Rule

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

The derivative of the numerator \( \cos(4x) + a \cos(2x) + b \) is:

\[ -4 \sin(4x) - 2a \sin(2x) \]

The derivative of the denominator \( x^4 \) is:

\[ 4x^3 \]

Thus, we apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{-4 \sin(4x) - 2a \sin(2x)}{4x^3} \]

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

Substituting \( x = 0 \) into the new limit gives:

\[ \frac{-4 \cdot 0 - 2a \cdot 0}{0} = \frac{0}{0} \]

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

### Step 4: Differentiate again

Differentiating the numerator again gives:

\[ -16 \cos(4x) - 2a \cdot 2 \cos(2x) = -16 \cos(4x) - 4a \cos(2x) \]

The derivative of the denominator \( 4x^3 \) is:

\[ 12x^2 \]

So we have:

\[ \lim_{x \to 0} \frac{-16 \cos(4x) - 4a \cos(2x)}{12x^2} \]

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

Substituting \( x = 0 \) gives:

\[ \frac{-16 \cdot 1 - 4a \cdot 1}{0} = \frac{-16 - 4a}{0} \]

For this limit to be finite, we need:

\[ -16 - 4a = 0 \]

### Step 6: Solve for \( a \)

Solving for \( a \):

\[ 4a = -16 \implies a = -4 \]

### Step 7: Substitute \( a \) back to find \( b \)

Now we substitute \( a = -4 \) back into the equation \( 1 + a + b = 0 \):

\[ 1 - 4 + b = 0 \implies b = 3 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:

\[ a = -4, \quad b = 3 \]