Let $I_n = \int_0^1 x^n\sqrt{1-x^2} dx$ then find the value of $\lim_{n \to \infty} \frac{I_n}{I_{n-2}}$.
Step-by-Step Solution
Key Concept: Integration by parts combined with recurrence relations determines the asymptotic behavior of sequences of integrals.
The integral $I_n = \int_0^1 x^n(1-x^2)^{3/2} dx$ is solved using integration by parts. Setting $u = x^n$ and $dv = (1-x^2)^{3/2} dx$, we obtain $I_n = 0 + \frac{n-1}{3}\int_0^1 x^{n-2}(1-x^2)^{1/2} dx$. This leads to the recurrence relation $3I_n + (n-1)I_{n-2} = (n-1)I_{n-2}$, which simplifies to show that $\lim_{n\to\infty} \frac{I_n}{I_{n-2}} = 1$.
Correct Answer: 1