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

Options

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

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Rewrite the limit
We can rewrite the limit as:

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

### Step 2: Combine the integrals
Using properties of integrals, we can combine the two integrals:

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

Thus, the limit becomes:

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

### Step 3: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0, leading to an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:

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

### Step 4: Differentiate the numerator and denominator
Using the Fundamental Theorem of Calculus, we differentiate the numerator:

\[ \frac{d}{dx} \left( \int_{y}^{x+y} e^{\sin^2 t} dt \right) = e^{\sin^2(x+y)} \cdot \frac{d}{dx}(x+y) = e^{\sin^2(x+y)}. \]

The derivative of the denominator \( x \) is simply \( 1 \).

### Step 5: Evaluate the limit
Now we have:

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

As \( x \to 0 \), this simplifies to:

\[ e^{\sin^2(y)}. \]

### Final Answer
Thus, the limit evaluates to:

\[ e^{\sin^2 y}. \]