`lim_(x rarr 1) (int_0^((x-1)^2) tcos tdt)/ ((x-1) sin (x-1))` is equal to

Options

A. `2`
B. `0`
C. `1`
D. Does not exist

Answer

To solve the limit

\[ \lim_{x \to 1} \frac{\int_0^{(x-1)^2} t \cos t \, dt}{(x-1) \sin (x-1)}, \]

we will follow these steps:

### Step 1: Check the form of the limit
Substituting \(x = 1\) into the limit gives:

- The numerator becomes \(\int_0^{(1-1)^2} t \cos t \, dt = \int_0^0 t \cos t \, dt = 0\).
- The denominator becomes \((1-1) \sin(1-1) = 0 \cdot \sin(0) = 0\).

Thus, we have the indeterminate form \(\frac{0}{0}\).

**Hint:** Always check the form of the limit before proceeding with L'Hôpital's rule.

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

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists.

### Step 3: Differentiate the numerator
Using the Leibniz rule for differentiation under the integral sign, we differentiate the numerator:

\[ \frac{d}{dx} \left( \int_0^{(x-1)^2} t \cos t \, dt \right) = (x-1)^2 \cos((x-1)^2) \cdot \frac{d}{dx}((x-1)^2) = (x-1)^2 \cos((x-1)^2) \cdot 2(x-1). \]

Thus, the derivative of the numerator is:

\[ 2(x-1)^3 \cos((x-1)^2). \]

### Step 4: Differentiate the denominator
Now we differentiate the denominator:

\[ \frac{d}{dx} \left( (x-1) \sin(x-1) \right) = \sin(x-1) + (x-1) \cos(x-1). \]

### Step 5: Rewrite the limit
Now we can rewrite our limit using the derivatives:

\[ \lim_{x \to 1} \frac{2(x-1)^3 \cos((x-1)^2)}{\sin(x-1) + (x-1) \cos(x-1)}. \]

### Step 6: Check the form again
Substituting \(x = 1\) again gives:

- The numerator becomes \(2(1-1)^3 \cos(0) = 0\).
- The denominator becomes \(\sin(0) + (1-1) \cos(0) = 0 + 0 = 0\).

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

### Step 7: Differentiate again
Differentiate the numerator again:

\[ \frac{d}{dx} \left( 2(x-1)^3 \cos((x-1)^2) \right) = 6(x-1)^2 \cos((x-1)^2) + 2(x-1)^3 \cdot (-\sin((x-1)^2)) \cdot 2(x-1). \]

The denominator differentiates to:

\[ \cos(x-1) + \cos(x-1) - (x-1) \sin(x-1) = 2\cos(x-1) - (x-1) \sin(x-1). \]

### Step 8: Evaluate the limit again
Substituting \(x = 1\) again into the new limit gives:

- The numerator becomes \(6(0)^2 \cos(0) + 2(0)^3 \cdot (-\sin(0)) \cdot 2(0) = 0\).
- The denominator becomes \(2\cos(0) - (0)\sin(0) = 2\).

Thus, we have:

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

### Final Answer
The limit is:

\[ \boxed{0}. \]