If `lim_(xrarr0)(x^3)/(sqrt(a+x)(bx-sinx))=1,agt0`, then `a+b` is equal to

Options

A. 36
B. 37
C. 38
D. 40

Explanation

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

\[ \lim_{x \to 0} \frac{x^3}{\sqrt{a+x} (bx - \sin x)} = 1 \]

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator \( x^3 \) and the denominator \( \sqrt{a+x} (bx - \sin x) \) approach 0, resulting in the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
Using L'Hôpital's Rule, we differentiate the numerator and denominator:

- The derivative of the numerator \( x^3 \) is \( 3x^2 \).
- The denominator is \( \sqrt{a+x} (bx - \sin x) \). We will need to use the product rule for differentiation.

Let \( u = \sqrt{a+x} \) and \( v = bx - \sin x \).

Using the product rule:

\[ \frac{d}{dx}(uv) = u'v + uv' \]

Calculating \( u' \) and \( v' \):

- \( u' = \frac{1}{2\sqrt{a+x}} \)
- \( v' = b - \cos x \)

Thus, the derivative of the denominator is:

\[ \frac{1}{2\sqrt{a+x}}(bx - \sin x) + \sqrt{a+x}(b - \cos x) \]

### Step 3: Apply L'Hôpital's Rule
Now we can write the limit again:

\[ \lim_{x \to 0} \frac{3x^2}{\frac{1}{2\sqrt{a}}(0 - 0) + \sqrt{a}(b - 1)} \]

As \( x \to 0 \), \( \sin x \to 0 \) and \( \cos x \to 1 \). Therefore, we simplify:

\[ = \lim_{x \to 0} \frac{3x^2}{\sqrt{a}(b - 1)} \]

### Step 4: Set the limit equal to 1
For the limit to equal 1, we need:

\[ \frac{3x^2}{\sqrt{a}(b - 1)} \to 1 \text{ as } x \to 0 \]

This implies:

\[ \sqrt{a}(b - 1) = 0 \]

Since \( a > 0 \), we must have \( b - 1 = 0 \), leading to:

\[ b = 1 \]

### Step 5: Substitute \( b \) back into the limit
Now we substitute \( b = 1 \) back into the limit:

\[ \lim_{x \to 0} \frac{3x^2}{\sqrt{a}(1 - 1)} \text{ is still } 0 \]

We need to differentiate again. The next derivatives will give:

\[ \lim_{x \to 0} \frac{6x}{\sqrt{a}(0 + \sin x)} = \lim_{x \to 0} \frac{6x}{\sqrt{a}x} = \frac{6}{\sqrt{a}} \]

Setting this equal to 1 gives:

\[ \frac{6}{\sqrt{a}} = 1 \implies \sqrt{a} = 6 \implies a = 36 \]

### Step 6: Calculate \( a + b \)
Now we have:

\[ a = 36, \quad b = 1 \]

Thus:

\[ a + b = 36 + 1 = 37 \]

### Final Answer
The value of \( a + b \) is:

\[ \boxed{37} \]