`lim_(x->0) 1/x [int_y ^a)e^(sin^2t) dt-int_(x+y) ^a)e^(sin^2t)dt]` is equal to (a) `e^(sin^(2)y)` (b) `sin2ye^(sin^(2)y` (c) `0` (d) none of these

Options

A. a. `e^(sin^(2)y)`
B. `sin2ye^(sin^(2)y`
C. `0`
D. none of these

Explanation

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

### Step 1: Rewrite the limit expression
We start with the limit expression:
\[ \lim_{x \to 0} \frac{1}{x} \left( \int_y^a e^{\sin^2 t} \, dt - \int_{x+y}^a e^{\sin^2 t} \, dt \right) \]

### Step 2: Apply the Fundamental Theorem of Calculus
Using the properties of definite integrals, we can rewrite the second integral:
\[ \int_{x+y}^a e^{\sin^2 t} \, dt = \int_y^a e^{\sin^2 t} \, dt - \int_y^{x+y} e^{\sin^2 t} \, dt \]
Thus, we can substitute this back into our limit:
\[ \lim_{x \to 0} \frac{1}{x} \left( \int_y^a e^{\sin^2 t} \, dt - \left( \int_y^a e^{\sin^2 t} \, dt - \int_y^{x+y} e^{\sin^2 t} \, dt \right) \right) \]
This simplifies to:
\[ \lim_{x \to 0} \frac{1}{x} \left( \int_y^{x+y} e^{\sin^2 t} \, dt \right) \]

### Step 3: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0, allowing us to apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\int_y^{x+y} e^{\sin^2 t} \, dt}{x} \]
Differentiating the numerator and denominator with respect to \( x \):
- The derivative of the denominator \( x \) is \( 1 \).
- For the numerator, we apply the Fundamental Theorem of Calculus:
\[ \frac{d}{dx} \int_y^{x+y} e^{\sin^2 t} \, dt = e^{\sin^2(x+y)} \cdot \frac{d}{dx}(x+y) = e^{\sin^2(x+y)} \]

### Step 4: Substitute and Evaluate the Limit
Now we have:
\[ \lim_{x \to 0} e^{\sin^2(x+y)} \]
As \( x \to 0 \), this approaches:
\[ e^{\sin^2(y)} \]

### Final Answer
Thus, the limit evaluates to:
\[ e^{\sin^2 y} \]

### Conclusion
The answer is \( e^{\sin^2 y} \), which corresponds to option (a).

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