`lim_(x to 0)1/x[int_(y)^(0) e^(sin^(2)t) dt - int_(x+y)^(0) e^(sin^(2)t) dt]` is equal to:

Options

A. `e^(sin^(2)y)`
B. `sin 2y e^(sin^(2)) y`
C. 0
D. none of these

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{1}{x} \left( \int_{y}^{0} e^{\sin^2 t} \, dt - \int_{x+y}^{0} e^{\sin^2 t} \, dt \right), \]

we can start by rewriting the expression inside the limit.

### Step 1: Rewrite the integrals
Using the property of definite integrals, we can express the difference of integrals as:

\[ \int_{y}^{0} e^{\sin^2 t} \, dt - \int_{x+y}^{0} e^{\sin^2 t} \, dt = \int_{y}^{x+y} e^{\sin^2 t} \, dt. \]

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin^2 t} \, dt. \]

### Step 2: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0, giving us the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that:

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

if the limit exists.

Let \( f(x) = \int_{y}^{x+y} e^{\sin^2 t} \, dt \) and \( g(x) = x \).

### Step 3: Differentiate the numerator and denominator
Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \):

\[ f'(x) = e^{\sin^2(x+y)} \cdot \frac{d}{dx}(x+y) = e^{\sin^2(x+y)} \cdot 1 = e^{\sin^2(x+y)}. \]

The derivative of the denominator \( g(x) \) is simply:

\[ g'(x) = 1. \]

### Step 4: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} e^{\sin^2(x+y)}. \]

### Step 5: Evaluate the limit
As \( x \to 0 \), we have:

\[ \sin^2(x+y) \to \sin^2(y). \]

Thus,

\[ \lim_{x \to 0} e^{\sin^2(x+y)} = e^{\sin^2(y)}. \]

### Final Result
Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{1}{x} \left( \int_{y}^{0} e^{\sin^2 t} \, dt - \int_{x+y}^{0} e^{\sin^2 t} \, dt \right) = e^{\sin^2(y)}. \]