$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

  • A
    $e^{1/2}$
  • B
    $0$
  • C
    $e^{-1}$
  • D
    $1$

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Similar Questions

$\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}} = $

દ્વિઘાત બહુપદી $p(x)$ ના બીજ $1$ અને $\alpha$ છે, જ્યારે દ્વિઘાત બહુપદી $q(x)$ ના બીજ $1$ અને $\beta$ છે. ધારો કે $\alpha$ અને $\beta$ એ $r(x) = p(x) + q(x)$ ના બીજ છે. તો $\lim_{x \to \infty} [\sqrt{p(x)} - \sqrt{q(x)}] = $

$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} = $

$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$ ની કિંમત શોધો.

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