$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}} = $

  • A
    $16$
  • B
    $24$
  • C
    $32$
  • D
    $8$

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यदि $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$ है,तो $\frac{\log _e \alpha}{1+\log _e \alpha}$ का मान ज्ञात कीजिए:

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