Q100 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

$\lim _{x \rightarrow \infty}\left(\frac{3 x^2-2 x+3}{3 x^2+x-2}\right)^{3 x-2}=$

  • A
    $-3$
  • B
    $e^{-1}$
  • Correct answer
    C
    $e^{-3}$
  • D
    $-1$

Solution

Since, $\lim _{x \rightarrow \infty}\left(\frac{3 x^2-2 x+3}{3 x^2+x-2}\right)^{3 x-2}$ $=\lim _{x \rightarrow \infty}\left(1+\frac{(-3 x+5)}{3 x^2+x-2}\right)^{3 x-2} \quad(\because 1^\infty$ forms) $=e^{\lim _{x \rightarrow \infty}\left(1+\frac{-3 x+5}{3 x^2+x-2}-1\right)(3 x-2)}$ $=e^{\lim _{x \rightarrow \infty}\left(\frac{-9 x^2+21 x-10}{3 x^2+x-2}\right)}=e^{\left(-\frac{9}{3}\right)}=e^{-3}$

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

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