Question:

Step 1: Recognize this as a Riemann sum and set up the limit. We need to convert the summation into an integral. Let $t = \frac{r}{n}$, so as $n \to \infty$, the sum becomes a Riemann sum with $\Delta t = \frac{1}{n}$. $$L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\left(\frac{(n+r)\pi}{4n}\right)}$$ Step 2: Simplify the argument of the sine function. Rewrite the argument inside the sine: $$\frac{(n+r)\pi}{4n} = \frac{n\pi}{4n} + \frac{r\pi}{4n} = \frac{\pi}{4} + \frac{r\pi}{4n}$$ Therefore: $$L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\left(\frac{\pi}{4} + \frac{r\pi}{4n}\right)}$$ Step 3: Convert the Riemann sum to a definite integral. As $n \to \infty$, with $t = \frac{r}{n}$, we have $\frac{\pi}{n} = \pi \cdot \frac{1}{n} = \pi \, dt$: $$L = \int_0^1 \frac{\pi}{\sin\left(\frac{\pi}{4} + \frac{\pi t}{4}\right)} dt = \int_0^1 \frac{\pi}{\sin\left(\frac{\pi(1+t)}{4}\right)} dt$$ Step 4: Use substitution to simplify the integral. Let $u = \frac{\pi(1+t)}{4}$, so $du = \frac{\pi}{4} dt$, which gives $dt = \frac{4}{\pi} du$. When $t = 0$: $u = \frac{\pi}{4}$ When $t = 1$: $u = \frac{\pi}{2}$ Substituting: $$L = \int_{\pi/4}^{\pi/2} \frac{\pi}{\sin u} \cdot \frac{4}{\pi} du = 4\int_{\pi/4}^{\pi/2} \csc(u) \, du$$ Step 5: Evaluate the integral using the antiderivative of cosecant. The antiderivative of $\csc(u)$ is $-\ln|\csc(u) + \cot(u)|$ or equivalently $\ln\left|\tan\left(\frac{u}{2}\right)\right|$. Using the second form: $$L = 4\left[\ln\left|\tan\left(\frac{u}{2}\right)\right|\right]_{\pi/4}^{\pi/2}$$ $$= 4\left[\ln\left(\tan\left(\frac{\pi}{4}\right)\right) - \ln\left(\tan\left(\frac{\pi}{8}\right)\right)\right]$$ Step 6: Evaluate the tangent values. We know that $\tan\left(\frac{\pi}{4}\right) = 1$, so: $$L = 4\left[\ln(1) - \ln\left(\tan\left(\frac{\pi}{8}\right)\right)\right] = 4\left[0 - \ln\left(\tan\left(\frac{\pi}{8}\right)\right)\right]$$ $$= -4\ln\left(\tan\left(\frac{\pi}{8}\right)\right)$$ Step 7: Use the known value of $\tan\left(\frac{\pi}{8}\right)$. The exact value is $\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1$. Therefore: $$L = -4\ln(\sqrt{2} - 1)$$ Step 8: Rationalize to express in the required form. Using the property $\ln(x) = -\ln\left(\frac{1}{x}\right)$: $$L = 4\ln\left(\frac{1}{\sqrt{2} - 1}\right)$$ Rationalize the denominator by multiplying by $\frac{\sqrt{2} + 1}{\sqrt{2} + 1}$: $$\frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$$ Therefore: $$L = 4\ln(\sqrt{2} + 1)$$ **Final Answer:** The value of the limit is $\boxed{4\ln(\sqrt{2}+1)}$, which corresponds to **Option 3**.