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

Answer: (D)

$\frac{1}{2} \tan 1$

Answer 2

$\displaystyle \lim_{n \to\infty} \left[ \frac{1}{n} \sec^{2} \frac{1}{n^{2}} + \frac{2}{n^{2}} \sec^{2} \frac{4}{n^{2} } + \frac{3}{n^{2}} \sec^{2} \frac{9}{n^{2}} + .... + \frac{1}{n} \sec^{2} 1 \right]$ 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 $x^2 = t$ ] $= \frac{1}{2} \left(\tan t\right)^{1}_{0} = \frac{1}{2} \tan 1$.