mathop {Lim}limits_{n to infty } frac{pi }{{6 | vedclass

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(A) The given expression is a Riemann sum of the form $\lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight) = \int_{0}^{1} f(x) dx

Vedclass · Accepted Answer

$\frac{{\sqrt 3 }}{3}$

Vedclass · Answer

(A) The given expression is a Riemann sum of the form $\lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight) = \int_{0}^{1} f(x) dx$. Let $f(x) = \frac{\pi}{6} \sec^2\left(\frac{\pi x}{6}ight)$. The expression can be written as $\sum_{r=1}^{n-1} \frac{\pi}{6n} \sec^2\left(\frac{r\pi}{6n}ight) + \frac{\pi}{6n} \cdot \frac{4}{3}$. As $n 	o \infty$,the term $\frac{\pi}{6n} \cdot \frac{4}{3} 	o 0$. Thus,the limit is $\int_{0}^{1} \frac{\pi}{6} \sec^2\left(\frac{\pi x}{6}ight) dx$. Let $u = \frac{\pi x}{6}$,then $du = \frac{\pi}{6} dx$. The integral becomes $\int_{0}^{\pi/6} \sec^2(u) du = [	an(u)]_{0}^{\pi/6}$. Evaluating this,we get $	an(\pi/6) - 	an(0) = \frac{1}{\sqrt{3}} - 0 = \frac{\sqrt{3}}{3}$.

$\mathop {Lim}\limits_{n \to \infty } \frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2 \cdot \frac{\pi }{{6n}}} \right) + \dots + {{\sec }^2}\left( {(n - 1)\frac{\pi }{{6n}}} \right) + \frac{4}{3}} \right]$ has the value equal to

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