$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+---+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

• A

  $\frac{\pi}{2}$
• B

  $\frac{\pi}{3}$
• C

  $\frac{\pi}{4}$
• Correct answer
  D

  $\frac{\pi}{6}$

Hence, the correct answer is option: 'D'.