`Suppose lim_(xrarr0) (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 `l` exists and is non- zero, then

Options

A. `b gt 1`
B. `0 lt b lt 1`
C. `b lt0`
D. `b=1`

Answer

To solve the given problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\int_{0}^{x} t^{2} \, dt}{(a + x^{r})^{1/p}} \Big/ (bx - \sin x) = l \]

where \( p \in \mathbb{N}, p \geq 2, a > 0, r > 0, \) and \( b \neq 0 \). We need to find the conditions under which \( l \) exists and is non-zero.

### Step 1: Evaluate the integral

First, we evaluate the integral in the numerator:

\[ \int_{0}^{x} t^{2} \, dt = \left[ \frac{t^{3}}{3} \right]_{0}^{x} = \frac{x^{3}}{3} \]

### Step 2: Substitute the integral into the limit

Now we substitute this result back into the limit:

\[ \lim_{x \to 0} \frac{\frac{x^{3}}{3}}{(a + x^{r})^{1/p}} \Big/ (bx - \sin x) \]

This simplifies to:

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

### Step 3: Analyze the denominator

Next, we analyze the denominator \( bx - \sin x \). Using the Taylor series expansion for \( \sin x \) around \( x = 0 \):

\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]

Thus,

\[ bx - \sin x \approx bx - \left(x - \frac{x^3}{6}\right) = (b-1)x + \frac{x^3}{6} \]

### Step 4: Substitute the approximation into the limit

Now we can rewrite our limit:

\[ \lim_{x \to 0} \frac{x^{3}}{3((b-1)x + \frac{x^{3}}{6})(a + x^{r})^{1/p}} \]

### Step 5: Determine the behavior as \( x \to 0 \)

As \( x \to 0 \), the term \( (b-1)x \) dominates \( \frac{x^{3}}{6} \) if \( b \neq 1 \). Thus, we can approximate:

\[ bx - \sin x \approx (b-1)x \]

### Step 6: Substitute this back into the limit

Substituting this back, we get:

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

### Step 7: Evaluate the limit

As \( x \to 0 \), \( a + x^{r} \to a \). Therefore, we have:

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

### Step 8: Condition for \( l \) to be non-zero

For \( l \) to exist and be non-zero, we need \( b - 1 = 0 \), which implies \( b = 1 \).

### Conclusion

Thus, if \( l \) exists and is non-zero, then:

\[ b = 1 \]