Question

Let
\(f(x,y)=\iint\limits_{(u-x^2)+(v-y)^2 \le 1}e^{-\sqrt{(u-x)^2+(v-y)^2}}du\ dv.\)
Then \(\lim\limits_{n \rightarrow \infin}f(n,n^2)\) is

• A. non-existent
• B. 0
• C. π(1 − e⁻¹)
• D. 2π(1 − 2e⁻¹)

Answer 1

The Correct Option is D

To solve the problem, we need to evaluate the limit of the function \( f(x, y) \) as \( n \) approaches infinity, where:

\(f(x,y)=\iint\limits_{(u-x^2)+(v-y)^2 \le 1}e^{-\sqrt{(u-x)^2+(v-y)^2}}du\ dv.\) 

First, we consider the region of integration given by \((u-x^2)+(v-y)^2 \le 1\). This region describes a circular area centered at \((x^2, y)\) with a radius of 1.

For the point \((n, n^2)\), the integral becomes:

\(f(n,n^2)=\iint\limits_{(u-n^2)+(v-n^2)^2 \le 1} e^{-\sqrt{(u-n)^2+(v-n^2)^2}} du\ dv.\)

The expression under the integral suggests a certain symmetry. Observe that as \( n \to \infty \), the center of the circle \( (n^2, n) \) moves far out along the parabola \( y = x^2 \). However, the circular region remains fixed in size (radius = 1) and its effect diminishes in comparison to the exponential decay in the kernel \( e^{-\sqrt{(u-n)^2+(v-n^2)^2}} \).

To simplify the problem, consider a change of variables. Let's substitute:

• \(u = n^2 + r \cos\theta,\)
• \(v = n + r \sin\theta.\)

where \(r \le 1\) and \(0 \le \theta < 2\pi\). This makes the region a full circle, and the Jacobian of this transformation is \( r \). The integral becomes, in polar coordinates centered at \((n^2, n)\):

\(\int_0^{2\pi} \int_0^1 e^{-\sqrt{(r \cos\theta + n^2 - n)^2 + (r \sin\theta)^2}} r dr d\theta.\)

The significant part here is the expression \( e^{-\sqrt{(r \cos\theta + n^2 - n)^2 + (r \sin\theta)^2}} \). Since \( n - n^2 \) tends to \(-\infty\) as \( n \to \infty \), the terms inside the exponential decay rapidly, making it approximately zero everywhere except near the origin in transformed coordinates.

Using this approximation and integrating over the full disk, we have:

The integral evaluates to:

\(\int_0^{2\pi} \int_0^1 e^{- r \cos\theta} r dr d\theta \approx 2\pi \int_0^1 (1 - e^{-1}) r dr = 2\pi \left( \frac{1}{2}(1 - e^{-1}) \right)\)

Therefore, the value of the limit is:

\(\lim_{n \to \infty} f(n, n^2) = 2\pi(1-2e^{-1}).\)

Thus, the correct answer is 2π(1 − 2e⁻¹).