The value of `lim_(xrarr0) (int_(0)^(x^2)sec^2t dt)/(x sin x) dx` , is

Options

A. 0
B. 3
C. 2
D. 1

Explanation

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

### Step 1: Identify the limit expression
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{\int_{0}^{x^2} \sec^2 t \, dt}{x \sin x} \]

### Step 2: Evaluate the integral in the numerator
First, we need to compute the integral:
\[ \int_{0}^{x^2} \sec^2 t \, dt \]
The integral of \(\sec^2 t\) is \(\tan t\). Therefore, we have:
\[ \int_{0}^{x^2} \sec^2 t \, dt = \tan(x^2) - \tan(0) = \tan(x^2) \]

### Step 3: Rewrite the limit expression
Now we can rewrite the limit expression as:
\[ \lim_{x \to 0} \frac{\tan(x^2)}{x \sin x} \]

### Step 4: Check the form of the limit
As \(x \to 0\), both the numerator and denominator approach 0, giving us the indeterminate form \(\frac{0}{0}\). Thus, we can apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule
We differentiate the numerator and denominator:
- The derivative of the numerator \(\tan(x^2)\) is:
\[ \frac{d}{dx} \tan(x^2) = \sec^2(x^2) \cdot 2x \]
- The derivative of the denominator \(x \sin x\) is:
\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x \]

### Step 6: Rewrite the limit with derivatives
Now we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2x \sec^2(x^2)}{\sin x + x \cos x} \]

### Step 7: Substitute \(x = 0\) into the new limit
Substituting \(x = 0\) gives:
- The numerator becomes:
\[ 2(0) \sec^2(0) = 0 \]
- The denominator becomes:
\[ \sin(0) + 0 \cdot \cos(0) = 0 \]
This is still an indeterminate form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again.

### Step 8: Differentiate again
Differentiate the numerator and denominator again:
- The derivative of the numerator \(2x \sec^2(x^2)\) is:
\[ 2 \sec^2(x^2) + 2x \cdot 2\sec^2(x^2) \tan(x^2) = 2\sec^2(x^2)(1 + 2x \tan(x^2)) \]
- The derivative of the denominator \(\sin x + x \cos x\) is:
\[ \cos x + \cos x - x \sin x = 2\cos x - x \sin x \]

### Step 9: Rewrite the limit again
Now we have:
\[ \lim_{x \to 0} \frac{2\sec^2(x^2)(1 + 2x \tan(x^2))}{2\cos x - x \sin x} \]

### Step 10: Substitute \(x = 0\) again
Substituting \(x = 0\):
- The numerator becomes:
\[ 2 \sec^2(0)(1 + 0) = 2 \cdot 1 \cdot 1 = 2 \]
- The denominator becomes:
\[ 2\cos(0) - 0 \cdot \sin(0) = 2 \cdot 1 - 0 = 2 \]

### Step 11: Final limit value
Thus, the limit evaluates to:
\[ \frac{2}{2} = 1 \]

### Conclusion
The value of the limit is:
\[ \boxed{1} \]