Mathematics

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


ANSWER

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


SOLUTION
$$\displaystyle \lim_{n\rightarrow \infty }\frac{1}{n^{2}}\sec ^{2}\frac{1}{n^{2}}+\frac{2}{n^{2}}\sec ^{2}\left ( \frac{4}{n^{2}} \right )+...+\frac{1}{n}\sec ^{2}1$$
   $$\displaystyle =\lim_{n\rightarrow \infty }\frac{1}{n^{2}}\sec ^{2}\frac{1}{n^{2}}+\frac{2}{n^{2}}\sec ^{2}\left ( \frac{4}{n^{2}} \right )+...+\frac{n}{n^{2}}\sec ^{2}\left ( \frac{n^{2}}{n^{2}} \right )$$
   $$\displaystyle =\lim_{n\rightarrow \infty }\sum_{r=1}^{r=n}\left ( \frac{r}{n^{2}} \right )\sec ^{2}\left ( \frac{r}{n} \right )^{2}=\lim_{n\rightarrow \infty }\sum_{r=1}^{r=n}\frac{1}{n}\left ( \frac{r}{n} \right )\sec ^{2}\left ( \frac{r}{n} \right )^{2}$$
   $$\displaystyle =\int_{0}^{1}x\sec ^{2}\left ( x^{2} \right )dx=\frac{1}{2}\tan 1$$
Hence, option 'C' is correct.
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Single Correct Hard Published on 17th 09, 2020
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