QJEEM25Ma0fec: JEE Main 2025 Mathematics - Class 12 JEEM25Ma0fec Solution

Question

Verified Solution • 2025–26 Session moderate

QJEEM25Ma0fec: JEE Main 2025 Mathematics - Class 12 JEEM25Ma0fec Solution

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

Multiple Choice Options

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

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Accepted Answer

$\begin{aligned} lim_{x \to \infty} \frac{(2 - \frac{3}{x} + \frac{5}{x^{2}}) {(1 - \frac{1}{3 x})}^{x / 2}}{(3 + \frac{5}{x} + \frac{4}{x^{2}}) {(1 + \frac{2}{3 x})}^{x / 2}} \\ = lim_{x \to \infty} \frac{2}{3} \cdot \frac{e^{\frac{x}{2} (1 - \frac{1}{3 x} - 1)}}{e^{\frac{x}{2} (1 + \frac{2}{3 x} - 1)}} \\ = \frac{2}{3} \cdot \frac{e^{- \frac{1}{6}}}{e^{1 / 3}} = \frac{2}{3} e^{- \frac{1}{2}} \end{aligned}$

QJEEM25Ma0fec: JEE Main 2025 Mathematics - Class 12 JEEM25Ma0fec Solution

Multiple Choice Options

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