Evaluate `lim_(xto0)(int_(0)^(x^(2))cos^(2)tdt)/(x sin x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\int_{0}^{x^2} \cos^2 t \, dt}{x \sin x}, \]

we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator and denominator approach 0. Thus, we have the indeterminate form \( \frac{0}{0} \).

**Hint:** Check if the limit results in an indeterminate form before proceeding with L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

The numerator is \( \int_{0}^{x^2} \cos^2 t \, dt \). By the Fundamental Theorem of Calculus and the chain rule, we differentiate it:

\[ \frac{d}{dx} \left( \int_{0}^{x^2} \cos^2 t \, dt \right) = \cos^2(x^2) \cdot \frac{d}{dx}(x^2) = \cos^2(x^2) \cdot 2x. \]

The denominator is \( x \sin x \). We differentiate it using the product rule:

\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x. \]

### Step 3: Rewrite the limit using derivatives
Now we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{2x \cos^2(x^2)}{\sin x + x \cos x}. \]

**Hint:** Remember to differentiate both the numerator and denominator separately when applying L'Hôpital's Rule.

### Step 4: Simplify the limit
Now we can substitute \( x = 0 \) into the new limit. First, we evaluate the numerator:

\[ 2x \cos^2(x^2) \to 2(0) \cos^2(0) = 0. \]

Next, we evaluate the denominator:

\[ \sin(0) + 0 \cdot \cos(0) = 0 + 0 = 0. \]

We again have the \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
We differentiate the numerator and denominator again:

- For the numerator \( 2x \cos^2(x^2) \):

Using the product rule:

\[ \frac{d}{dx}(2x \cos^2(x^2)) = 2 \cos^2(x^2) + 2x \cdot 2\cos(x^2)(-\sin(x^2) \cdot 2x) = 2 \cos^2(x^2) - 8x^2 \cos(x^2) \sin(x^2). \]

- For the denominator \( \sin x + x \cos x \):

Using the product rule again:

\[ \frac{d}{dx}(\sin x + x \cos x) = \cos x + (\cos x - x \sin x) = 2\cos x - x \sin x. \]

### Step 6: Rewrite the limit again
Now we have:

\[ \lim_{x \to 0} \frac{2 \cos^2(x^2) - 8x^2 \cos(x^2) \sin(x^2)}{2\cos x - x \sin x}. \]

### Step 7: Evaluate the limit
Substituting \( x = 0 \):

- The numerator becomes:

\[ 2 \cos^2(0) - 8(0)^2 \cos(0) \sin(0) = 2(1) - 0 = 2. \]

- The denominator becomes:

\[ 2\cos(0) - 0 \cdot \sin(0) = 2(1) - 0 = 2. \]

Thus, we have:

\[ \lim_{x \to 0} \frac{2}{2} = 1. \]

### Final Answer
The limit is

\[ \boxed{1}. \]