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

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

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Similar Questions

$\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ का मान क्या है?

यदि $a$ और $b$ धनात्मक पूर्णांक हैं जैसे कि $b > a$,तो $\lim_{n \to \infty} \left[ \frac{1}{na} + \frac{1}{na + 1} + \frac{1}{na + 2} + \dots + \frac{1}{nb} \right] = $

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ का मान है:

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1^2}{1^3 + n^3} + \frac{2^2}{2^3 + n^3} + \dots + \frac{n^2}{n^3 + n^3} \right)$ का मान ज्ञात कीजिए।

Difficult
View Solution

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

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