Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 23 January, Evening Shift — Question 20

lim⁡x→∞(2x2−3x+5)(3x−1)x2(3x2+5x+4)(3x+2)x is equal to:\lim_{x\rightarrow\infty}\frac{(2x^{2}-3x+5)(3x-1)^{\frac{x}{2}}}{(3x^{2}+5x+4)\sqrt{(3x+2)^{x}}} \text{ is equal to:}
  1. Option A:

    23e\frac{2}{\sqrt{3 e}}

  2. Option B:

    2e3\frac{2 e}{\sqrt{3}}

  3. Option C:

    2e3\frac{2 e}{3}

  4. Option D:

    23e\frac{2}{3 \sqrt{\mathrm{e}}}

    Correct

Answer: D

Step-by-step solution

lim⁡x→∞(2−3x+5x2)(1−13x)x/2(3+5x+4x2)(1+23x)x/2\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→∞23⋅ex2(1−13x−1)ex2(1+23x−1)=23⋅e−16e1/3=23e−12\begin{aligned} & =\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 \mathrm{e}^{-\frac{1}{6}} \mathrm{e}^{1 / 3}=\frac{2}{3} \mathrm{e}^{-\frac{1}{2}} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Indeterminate forms & its solving methods