Suppose `lim_(x to 0)(int_(0)^(x)(t^(2) dt)/((a+t^(r))^(1//p)))/(bx- sinx)=l`,
`p in N, p ge 2,a gt gt 0,rgt 0 and b ne 0`
If `p=2 and a=9 and l` exists , then the value of `l` is equal to

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. `(1)/(3)`
D. `(7)/(9)`

Answer

To solve the given limit problem, we will follow these steps systematically:

### Step 1: Set up the limit expression
We start with the limit expression given in the problem:
\[ \lim_{x \to 0} \frac{\int_{0}^{x} t^2 \, dt}{(9 + t^r)^{1/2}} \div (bx - \sin x) \]
Given that \( p = 2 \) and \( a = 9 \), we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\int_{0}^{x} t^2 \, dt}{(9 + t^r)^{1/2}} \div (bx - \sin x) \]

### Step 2: Evaluate the integral
The integral \(\int_{0}^{x} t^2 \, dt\) can be computed as:
\[ \int_{0}^{x} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{x} = \frac{x^3}{3} \]

### Step 3: Substitute the integral back into the limit
Substituting the result of the integral back into the limit, we have:
\[ \lim_{x \to 0} \frac{\frac{x^3}{3}}{(9 + x^r)^{1/2}} \div (bx - \sin x) \]

### Step 4: Simplify the denominator
Using the Taylor series expansion for \(\sin x\) around \(x = 0\):
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]
Thus, we can write:
\[ bx - \sin x \approx (b - 1)x + \frac{x^3}{6} \]

### Step 5: Rewrite the limit
Now, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\frac{x^3}{3}}{(9 + x^r)^{1/2} \cdot ((b - 1)x + \frac{x^3}{6})} \]

### Step 6: Evaluate the limit
As \(x \to 0\), the term \((9 + x^r)^{1/2}\) approaches \(3\) (since \(9 + 0 = 9\)), and we have:
\[ \lim_{x \to 0} \frac{\frac{x^3}{3}}{3 \cdot ((b - 1)x + \frac{x^3}{6})} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{x^3}{9((b - 1)x + \frac{x^3}{6})} \]

### Step 7: Apply L'Hôpital's Rule
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- The derivative of the numerator \(x^3\) is \(3x^2\).
- The derivative of the denominator \(9((b - 1)x + \frac{x^3}{6})\) is \(9(b - 1) + \frac{9}{2}x^2\).

Thus, we have:
\[ \lim_{x \to 0} \frac{3x^2}{9(b - 1) + \frac{9}{2}x^2} \]

### Step 8: Evaluate the limit again
As \(x \to 0\):
\[ \lim_{x \to 0} \frac{3x^2}{9(b - 1)} = 0 \]
This indicates that \(b - 1\) must equal \(0\) for the limit \(L\) to exist, which gives \(b = 1\).

### Step 9: Final evaluation
Substituting \(b = 1\) back into the limit, we can evaluate:
\[ \lim_{x \to 0} \frac{x^3}{9 \cdot \frac{x^3}{6}} = \frac{6}{9} = \frac{2}{3} \]

### Conclusion
Thus, the value of \(l\) is:
\[ \boxed{\frac{2}{3}} \]