Question

$\displaystyle \lim_{n \to\infty} \left[\frac{1}{n^{2}} \sec^{2} \frac{1}{n^{2}} + \frac{2}{n^{2}} \sec^{2} \frac{4}{n^{2}}.......... + \frac{1}{n}\sec^{2} 1 \right] $ equals

• A. $\frac{1}{2} \sec 1 $
• B. $\frac{1}{2} cosec\, 1 $
• C. $\tan 1 $
• D. $\frac{1}{2} \tan 1 $

Answer 1

The Correct Option is D

$\displaystyle \lim_{n \to\infty} \left[ \frac{1}{n} \sec^{2} \frac{1}{n^{2}} + \frac{2}{n^{2}} \sec^{2} \frac{4}{n^{2} } + \frac{3}{n^{2}} \sec^{2} \frac{9}{n^{2}} + .... + \frac{1}{n} \sec^{2} 1 \right] $ is equal to $\displaystyle \lim_{n \to\infty} \frac{r}{n^{2} } \sec^{2} \frac{r^{2}}{n^{2}} = \displaystyle \lim_{n \to\infty} \frac{1}{n} . \frac{r}{n} \sec^{2} \frac{r^{2}}{n^{2}}$ $\Rightarrow $ Given limit is equal to value of integral $\int\limits^{1}_{0} x \sec^{2} x^{2}dx$ or $ \frac{1}{2} \int\limits^{1}_{0} 2x \sec x^{2} dx = \frac{1}{2} \int\limits^{1}_{0} \sec^{2} tdt$ [put $x^2 = t $ ] $ = \frac{1}{2} \left(\tan t\right)^{1}_{0} = \frac{1}{2} \tan 1 $.