$\lim _{\mathrm{n} \rightarrow \infty}\left[\frac{1}{\mathrm{n}^{2}} \sec ^{2} \frac{1}{\mathrm{n}^{2}}+\frac{2}{\mathrm{n}^{2}} \sec ^{2} \frac{4}{\mathrm{n}^{2}}+\ldots .+\frac{1}{\mathrm{n}^{2}} \sec ^{2} 1\right]$ equals

(B) $\frac{1}{2} \operatorname{cosec} 1$

JEE Main · 2005 · 08 May 2005 · Offline · Limits

lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots . + \frac{1}{n ^{2}} sec^{2} 1]

equals

Answer: (D)

The general term is

\frac{r}{n ^{2}} sec^{2} \frac{r ^{2}}{n ^{2}} = \frac{1}{n} \cdot \frac{r}{n} sec^{2} (\frac{r}{n})^{2}

, which is the Riemann sum for the function

x sec^{2} (x^{2})

[0, 1]

. As

n \to \infty

the sum tends to

\int_{0}^{1} x sec^{2} (x^{2}) d x

. Substituting

t = x^{2}

d t = 2 x d x

, this is

\frac{1}{2} \int_{0}^{1} sec^{2} t d t = \frac{1}{2} [tan t]_{0}^{1} = \frac{1}{2} tan 1

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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