Q102 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics

$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}=$

  • A
    $\frac{3}{2}$
  • Correct answer
    B
    $2 \sqrt{2}$
  • C
    3
  • D
    $\sqrt{2}$

Solution

Let $L=\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{|x|-1}}$ $$ L=\lim _{x \rightarrow \infty}\left(\frac{2+\frac{3}{x}+\frac{4}{x^2}}{1-\frac{3}{x}+\frac{5}{x^2}}\right)^{\frac{3+\frac{1}{|x|}}{2-\frac{1}{|x|}}}=\left(\frac{2}{1}\right)^{3 / 2}=2 \sqrt{2} $$

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

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