`lim_(xrarroo)((int_(0)^(x)e^(t^(2))dt)^(2))/(int_(0)^(x)e^(2t^(2))dt)` is equal to

Options

A. 1
B. 0
C. `-1`
D. none of these

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{\left(\int_{0}^{x} e^{t^2} dt\right)^2}{\int_{0}^{x} e^{2t^2} dt} \]

we will evaluate the integrals in the numerator and denominator separately.

### Step 1: Evaluate the numerator

The numerator is

\[ \left(\int_{0}^{x} e^{t^2} dt\right)^2 \]

Using L'Hôpital's Rule, we first find the derivative of the integral. By the Fundamental Theorem of Calculus, we have:

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

Thus, the derivative of the square is:

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

### Step 2: Evaluate the denominator

The denominator is

\[ \int_{0}^{x} e^{2t^2} dt \]

Again, using the Fundamental Theorem of Calculus, we find:

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

### Step 3: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule since both the numerator and denominator approach infinity as \( x \to \infty \):

\[ \lim_{x \to \infty} \frac{2\left(\int_{0}^{x} e^{t^2} dt\right)e^{x^2}}{e^{2x^2}} \]

### Step 4: Simplify the limit

This simplifies to:

\[ \lim_{x \to \infty} \frac{2\left(\int_{0}^{x} e^{t^2} dt\right)}{e^{x^2}} \]

### Step 5: Analyze the behavior of the integral

As \( x \to \infty \), the integral \( \int_{0}^{x} e^{t^2} dt \) grows very quickly, but we need to understand its growth relative to \( e^{x^2} \).

Using the asymptotic behavior of the integral, we can say that:

\[ \int_{0}^{x} e^{t^2} dt \sim \frac{e^{x^2}}{2x} \]

Thus, we can substitute this approximation into our limit:

\[ \lim_{x \to \infty} \frac{2 \cdot \frac{e^{x^2}}{2x}}{e^{x^2}} = \lim_{x \to \infty} \frac{1}{x} \]

### Step 6: Final limit

As \( x \to \infty \), \( \frac{1}{x} \to 0 \).

Thus, the final answer is:

\[ \lim_{x \to \infty} \frac{\left(\int_{0}^{x} e^{t^2} dt\right)^2}{\int_{0}^{x} e^{2t^2} dt} = 0 \]