Medium MCQ +4 / -1 PYQ · JEE Mains 2025

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

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

Solution

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

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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