Calculate the reciprocal of the limit `lim_(x->oo) int_0^x xe^(t^2-x^2) dt`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to calculate the reciprocal of the limit

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

Let's denote this limit as \( L \):

\[ L = \lim_{x \to \infty} \int_0^x x e^{t^2 - x^2} \, dt. \]

### Step 1: Rewrite the Integral

We can rewrite the integral as:

\[ L = \lim_{x \to \infty} x \int_0^x e^{t^2 - x^2} \, dt. \]

### Step 2: Analyze the Exponential Term

Notice that as \( x \to \infty \), the term \( e^{t^2 - x^2} \) will tend to zero for any fixed \( t \) because \( -x^2 \) will dominate \( t^2 \). Therefore, we can analyze the behavior of the integral.

### Step 3: Change of Variables

To better understand the limit, we can perform a change of variables. Let \( u = \frac{t}{x} \), then \( t = ux \) and \( dt = x \, du \). The limits of integration change from \( t = 0 \) to \( t = x \) into \( u = 0 \) to \( u = 1 \). Thus, we have:

\[ L = \lim_{x \to \infty} x \int_0^1 e^{(ux)^2 - x^2} x \, du = \lim_{x \to \infty} x^2 \int_0^1 e^{x^2(u^2 - 1)} \, du. \]

### Step 4: Analyze the Integral

Now, we analyze the integral:

\[ \int_0^1 e^{x^2(u^2 - 1)} \, du. \]

As \( x \to \infty \), for \( u < 1 \), \( u^2 - 1 < 0 \), thus \( e^{x^2(u^2 - 1)} \to 0 \). The only contribution to the integral comes from \( u \) very close to 1.

### Step 5: Dominated Convergence Theorem

Using the Dominated Convergence Theorem, we can conclude that:

\[ \int_0^1 e^{x^2(u^2 - 1)} \, du \to 0 \text{ as } x \to \infty. \]

### Step 6: Conclusion for \( L \)

Thus,

\[ L = \lim_{x \to \infty} x^2 \cdot 0 = 0. \]

### Step 7: Find the Reciprocal

Since \( L = 0 \), the reciprocal is:

\[ \frac{1}{L} = \frac{1}{0} \text{ which is undefined.} \]

However, if we consider the limit approaching infinity, we can say:

\[ \frac{1}{L} \to 0. \]

### Final Answer

The reciprocal of the limit is:

\[ \text{Reciprocal} = 0. \]