Q32 / 75 · JEE Main 2025 (23 Jan Shift 2) · Mathematics

$\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}$ is equal to :

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

Solution

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{x / 2}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} \\ & =\lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{5}{x^2}\right)(3 x)^{\frac{x}{2}}}{x^2\left(3+\frac{5}{x}+\frac{4}{x^2}\right)(3 x)^{\frac{x}{2}}} \cdot \frac{\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}}{\left(1+\frac{2}{3 x}\right)^{\frac{x}{2}}} \\ & \lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}=e^{\lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}-1\right) \times \frac{x}{2}}=e^{\frac{-1}{6}} \\ & \lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}\right)=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}-1\right) \times \frac{x}{2}} \\ & =e^{\frac{1}{3}} \end{aligned}$ So, $\frac{2}{3} \times \frac{e^{\frac{-1}{6}}}{e^{\frac{1}{3}}}=\frac{2}{3} \times \frac{1}{e^{\frac{1}{3}+\frac{1}{6}}}=\frac{2}{3 \sqrt{e}}$

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

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