दिया गया है कि $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. यदि $f: R \rightarrow R$ को $f(x)=x^2+2$ द्वारा परिभाषित किया गया है, तो $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$

  • A
    $55$
  • B
    $57$
  • C
    $104$
  • D
    $7$

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$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots \left(1+\frac{n^3}{n^3}\right)^{\frac{n^2}{n^3}}\right]=$

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

$a \in R, |a| > 1$ के लिए,मान लीजिए $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. तो $a$ का/के संभावित मान है/हैं:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ का मान ज्ञात कीजिए।

$\lim _{n \rightarrow \infty} \frac{1}{n} \left\{ \sec ^{2} \frac{\pi}{4 n} + \sec ^{2} \frac{2 \pi}{4 n} + \ldots + \sec ^{2} \frac{n \pi}{4 n} \right\}$ का मान ज्ञात कीजिए।

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