`lim_(xrarr0) (int_(0)^(x)(t^(2))/(sqrt(a+t))dt)/(x-sinx)=1(agt0)`. Then the value of a is

Options

A. `1//2`
B. `1//4`
C. 2
D. 4

Explanation

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

### Step 1: Identify the limit and the form
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{\int_0^x \frac{t^2}{\sqrt{a+t}} \, dt}{x - \sin x} \]
When we substitute \(x = 0\), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \).

**Hint:** Check if the limit results in an indeterminate form to decide if L'Hôpital's Rule is applicable.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we apply L'Hôpital's Rule, which states that if \(\lim_{x \to c} f(x) = 0\) and \(\lim_{x \to c} g(x) = 0\), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
We need to differentiate the numerator and denominator.

**Hint:** Differentiate the numerator and denominator separately.

### Step 3: Differentiate the numerator
Using the Fundamental Theorem of Calculus, the derivative of the numerator is:
\[ \frac{d}{dx} \left( \int_0^x \frac{t^2}{\sqrt{a+t}} \, dt \right) = \frac{x^2}{\sqrt{a+x}} \]

**Hint:** Remember to use the Fundamental Theorem of Calculus for differentiating integrals.

### Step 4: Differentiate the denominator
The derivative of the denominator \(x - \sin x\) is:
\[ \frac{d}{dx}(x - \sin x) = 1 - \cos x \]

**Hint:** Use basic differentiation rules for trigonometric functions.

### Step 5: Rewrite the limit
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\frac{x^2}{\sqrt{a+x}}}{1 - \cos x} \]
Again, substituting \(x = 0\) gives us \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

**Hint:** If you still have an indeterminate form, apply L'Hôpital's Rule again.

### Step 6: Differentiate again
Differentiate the numerator:
\[ \frac{d}{dx} \left( \frac{x^2}{\sqrt{a+x}} \right) = \frac{2x\sqrt{a+x} - \frac{x^2}{2\sqrt{a+x}}}{a+x} = \frac{2x(a+x) - \frac{x^2}{2}}{(a+x)\sqrt{a+x}} \]
And differentiate the denominator:
\[ \frac{d}{dx}(1 - \cos x) = \sin x \]

### Step 7: Rewrite the limit again
Now we have:
\[ \lim_{x \to 0} \frac{\frac{2x(a+x) - \frac{x^2}{2}}{(a+x)\sqrt{a+x}}}{\sin x} \]

### Step 8: Substitute \(x = 0\)
Substituting \(x = 0\) gives:
\[ \frac{0}{0} \text{ (still indeterminate)} \]
We can simplify the expression further or apply L'Hôpital's Rule again.

### Step 9: Evaluate the limit
After simplification and substituting \(x = 0\), we find:
\[ \lim_{x \to 0} \frac{2\sqrt{a}}{a} = \frac{2}{\sqrt{a}} \]
Setting this equal to 1 (as given in the problem):
\[ \frac{2}{\sqrt{a}} = 1 \implies \sqrt{a} = 2 \implies a = 4 \]

### Conclusion
The value of \(a\) is:
\[ \boxed{4} \]