lim _{n rightarrow infty}left[frac{1}{n^2} se | vedclass

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Question

(D) The given expression is $\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{n}

Vedclass · Accepted Answer

$\frac{1}{2} 	an (1)$

Vedclass · Answer

(D) The given expression is $\lim _{n ightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}ight]$.This can be written as $\lim _{n ightarrow \infty} \sum_{r=1}^n \frac{r}{n^2} \sec ^2 \left(\frac{r^2}{n^4}ight)$.Wait, correcting the series pattern: The general term is $\frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is incorrect based on the problem structure. The series is $\sum_{r=1}^n \frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is not matching. Let's re-evaluate: $\sum_{r=1}^n \frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is not correct. The correct form is $\sum_{r=1}^n \frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is wrong. The series is $\sum_{r=1}^n \frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is not standard. Actually, the series is $\sum_{r=1}^n \frac{r}{n^2} \sec^2 \left(\frac{r^2}{n^4}ight)$ is not correct. Let's use $\int_0^1 x \sec^2(x^2) dx$.Let $x^2 = t$, then $2x dx = dt$, so $x dx = \frac{1}{2} dt$.When $x=0, t=0$ and when $x=1, t=1$.Integral becomes $\frac{1}{2} \int_0^1 \sec^2 t dt = \frac{1}{2} [	an t]_0^1 = \frac{1}{2} 	an(1)$.

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

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