Question:

Step 1: Define the integral we need to evaluate. We need to find $L = \lim_{n\to\infty} \sqrt{n} \int_0^1 \dfrac{dx}{(1+x^2)^n}$. Let us denote $I_n = \int_0^1 \dfrac{dx}{(1+x^2)^n}$ so that $L = \lim_{n\to\infty} \sqrt{n} \cdot I_n$. Step 2: Analyze the behavior of the integrand for large $n$. For large values of $n$, the integrand $(1+x^2)^{-n}$ becomes increasingly concentrated near $x=0$, since the function decays rapidly as $x$ moves away from zero. This suggests we should use a substitution to capture this concentration. Step 3: Apply a substitution to rescale the integral. To analyze the behavior near $x=0$, we substitute $x = \dfrac{t}{\sqrt{n}}$, which gives $dx = \dfrac{dt}{\sqrt{n}}$. The limits of integration change: when $x=0$, we have $t=0$; when $x=1$, we have $t=\sqrt{n}$. Therefore: $$I_n = \int_0^{\sqrt{n}} \dfrac{1}{(1+t^2/n)^n} \cdot \dfrac{dt}{\sqrt{n}} = \dfrac{1}{\sqrt{n}} \int_0^{\sqrt{n}} \dfrac{dt}{(1+t^2/n)^n}$$ Step 4: Evaluate the limit of the integrand using exponential approximation. As $n \to \infty$, we use the standard limit: $$(1+t^2/n)^n \to e^{t^2}$$ This is because $\lim_{n\to\infty} (1+u/n)^n = e^u$ for any fixed $u$, and here $u = t^2$. Step 5: Apply the limit to find $\sqrt{n} \cdot I_n$. Multiplying both sides of the expression for $I_n$ by $\sqrt{n}$: $$\sqrt{n} \cdot I_n = \int_0^{\sqrt{n}} \dfrac{dt}{(1+t^2/n)^n}$$ As $n \to \infty$, the upper limit $\sqrt{n} \to \infty$ and the integrand converges to $e^{-t^2}$. Therefore: $$L = \lim_{n\to\infty} \sqrt{n} \cdot I_n = \int_0^{\infty} e^{-t^2}\, dt$$ Step 6: Evaluate the Gaussian integral. The integral $\int_0^{\infty} e^{-t^2}\, dt$ is a standard Gaussian integral: $$\int_0^{\infty} e^{-t^2}\, dt = \dfrac{\sqrt{\pi}}{2}$$ Therefore: $$L = \dfrac{\sqrt{\pi}}{2}$$ Step 7: Calculate the numerical value and verify the answer. Computing the numerical value: $$L = \dfrac{\sqrt{\pi}}{2} \approx \dfrac{1.7725}{2} \approx 0.886$$ We check which option this satisfies: - Option 1: $\dfrac{1}{2} < L < 2$ means $0.5 < 0.886 < 2$ ✓ Since $0.886$ lies strictly between $0.5$ and $2$, the correct answer is **Option 1: $\dfrac{1}{2} < L < 2$**.