Q115 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
$\operatorname{Lim}_{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=$
- Correct answerA1
- B0
- C4
- D3
Solution
We have,
$\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]$
Here, $r$ th term $=\sin \left(\frac{r \pi}{2 n}\right)$
$\therefore$ The given limit, $\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \sum_{r=1}^n \sin \left(\frac{r \pi}{2 n}\right)$
Put $\frac{r \pi}{2 n} \rightarrow x, \frac{\pi}{2 \pi} \rightarrow d x, \quad \lim _{n \rightarrow \infty} \Sigma \rightarrow \int$
$x_{\min }=\lim _{n \rightarrow \infty} \frac{r_{\min }}{2 n / \pi}=0$
$x_{\max }=\lim _{n \rightarrow \infty} \frac{r_{\max }}{2 n / \pi}=\frac{\pi}{2}=\int_0^{\pi / 2} \sin x d x$
$=(-\cos x)_0^{\pi / 2}=-(\cos x)_0^{\pi / 2}$
$=-\left[\cos \frac{\pi}{2}-\cos 0\right]=-[0-1]=1$
Hence, the correct answer is option: 'A'.