_ n arrow ∞ [(1)/(n^2) ^2 (1)/(n^2)+(2)/(n^2) ^2 (4)/(n^2)+ .+(1)/(n^2) ^2 1] equals

Correct answer: D. (1)/(2) 1 _ n arrow ∞ [(1)/(n^2) ^2 (1)/(n^2)+(2)/(n^2) ^2 (4)/(n^2)+(3)/(n^2) ^2 (9)/(n^2)+ .+(1)/(n) ^2 1] is equal to _ n arrow ∞ (r)/(n^2) ^2 (r^2)/(n^2)= _ n arrow ∞ (1)/(n) · (r)/(n) ^2 (r^2)/(n^2) Given limit is equal to value of integral ∫_0^1 x ^2 x^2 d x or (1)/(2) ∫_0^1 2 x x^2 d x=(1)/(2) ∫_0^1 ^2 t d t [put .x^2=t] =(1)/(2)( t)_0^1=(1)/(2) 1 .

JEE Main 2005

Exam: JEE Main
Year: 2005
Subject: Mathematics
Chapter: Limits

JEE Main 2005 · Mathematics · Limits

_ n arrow ∞ [(1)/(n^2) ^2 (1)/(n^2)+(2)/(n^2) ^2 (4)/(n^2)+...

\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}+\ldots .+\frac{1}{n^2} \sec ^2 1\right]

equals

Options

A $\frac{1}{2} \sec 1$
B $\frac{1}{2} \operatorname{cosec} 1$
C $\tan 1$
D $\frac{1}{2} \tan 1$

Correct answer

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

Step-by-step solution

\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{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots .+\frac{1}{n} \sec ^2 1\right]

is equal to

\lim _{n \rightarrow \infty} \frac{r}{n^2} \sec ^2 \frac{r^2}{n^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{r}{n} \sec ^2 \frac{r^2}{n^2}

\Rightarrow

Given limit is equal to value of integral

\int_0^1 x \sec ^2 x^2 d x

\frac{1}{2} \int_0^1 2 x \sec x^2 d x=\frac{1}{2} \int_0^1 \sec ^2 t d t

[put

\left.x^2=t\right]

=\frac{1}{2}(\tan t)_0^1=\frac{1}{2} \tan 1

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