`lim_(x to oo) (int_(0)^(2x) x e^(x^(2)))/e^(4x^(2))` equals

Options

A. 0
B. `oo`
C. 2
D. `(1)/(2)`

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{\int_{0}^{2x} x e^{x^2} \, dt}{e^{4x^2}}, \]

we will first analyze the integral in the numerator.

### Step 1: Simplify the Integral

The integral

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

can be simplified because \(x e^{x^2}\) is treated as a constant with respect to \(t\). Therefore, we can factor \(x e^{x^2}\) out of the integral:

\[ \int_{0}^{2x} x e^{x^2} \, dt = x e^{x^2} \int_{0}^{2x} dt = x e^{x^2} \cdot (2x - 0) = 2x^2 e^{x^2}. \]

### Step 2: Substitute Back into the Limit

Now we substitute this result back into the limit:

\[ \lim_{x \to \infty} \frac{2x^2 e^{x^2}}{e^{4x^2}}. \]

### Step 3: Simplify the Expression

Next, we simplify the fraction:

\[ \frac{2x^2 e^{x^2}}{e^{4x^2}} = 2x^2 e^{x^2 - 4x^2} = 2x^2 e^{-3x^2}. \]

### Step 4: Evaluate the Limit

Now we need to evaluate the limit:

\[ \lim_{x \to \infty} 2x^2 e^{-3x^2}. \]

As \(x \to \infty\), the term \(e^{-3x^2}\) approaches \(0\) much faster than \(2x^2\) approaches \(\infty\). Therefore, the limit evaluates to:

\[ \lim_{x \to \infty} 2x^2 e^{-3x^2} = 0. \]

### Final Answer

Thus, we conclude that:

\[ \lim_{x \to \infty} \frac{\int_{0}^{2x} x e^{x^2} \, dt}{e^{4x^2}} = 0. \]

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