`lim_(x to oo)(int_0^(2x) te^(t^(2))dt)/(e^(4x^(2)))`equals

Options

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

Explanation

To solve the limit

\[ \lim_{x \to \infty} \frac{\int_0^{2x} t e^{t^2} dt}{e^{4x^2}}, \]

we can apply L'Hôpital's Rule, which is useful when we encounter an indeterminate form like \(\frac{\infty}{\infty}\).

### Step 1: Identify the functions
Let
\[ f(x) = \int_0^{2x} t e^{t^2} dt \quad \text{and} \quad g(x) = e^{4x^2}. \]

### Step 2: Differentiate the numerator and denominator
We need to differentiate \(f(x)\) and \(g(x)\) with respect to \(x\).

Using the Fundamental Theorem of Calculus and the chain rule, we have:
\[ f'(x) = \frac{d}{dx} \left( \int_0^{2x} t e^{t^2} dt \right) = 2x e^{(2x)^2} = 2x e^{4x^2}. \]

For the denominator:
\[ g'(x) = \frac{d}{dx} (e^{4x^2}) = 8x e^{4x^2}. \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{2x e^{4x^2}}{8x e^{4x^2}}. \]

### Step 4: Simplify the expression
We can simplify the fraction:
\[ \frac{2x e^{4x^2}}{8x e^{4x^2}} = \frac{2}{8} = \frac{1}{4}. \]

### Step 5: Evaluate the limit
Thus, we find:
\[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \frac{1}{4}. \]

### Conclusion
Therefore, the limit is:
\[ \lim_{x \to \infty} \frac{\int_0^{2x} t e^{t^2} dt}{e^{4x^2}} = \frac{1}{4}. \]