_ x arrow ∞ (2 x^2-3 x+5)(3 x-1)^ (x)/(2) (3 x^2+5 x+4) (3 x+2)^x is equal to :

Correct answer: C. 2 3 e aligned & _ x arrow ∞ (2 x^2-3 x+5)(3 x-1)^ x / 2 (3 x^2+5 x+4) (3 x+2)^x & = _ x arrow ∞ x^2(2-(3)/(x)+(5)/(x^2))(3 x)^ (x)/(2) x^2(3+(5)/(x)+(4)/(x^2))(3 x)^ (x)/(2) · (1-(1)/(3 x))^ (x)/(2) (1+(2)/(3 x))^ (x)/(2) & _ x arrow ∞ (1-(1)/(3 x))^ (x)/(2) =e^ _ x arrow ∞ (1-(1)/(3 x)-1) × (x)/(2) =e^ (-1)/(6) & _ x arrow ∞ (1+(2)/(3 x))=e^ _ x arrow ∞ (1+(2)/(3 x)-1) × (x)/(2) & =e^ (1)/(3) aligned So, (2)/(3) × e^ (-1)/(6) e^ (1)/(3) =(2)/(3) × 1 e^ (1)/(3)+(1)/(6) = 2 3 e

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Limits

_ x arrow ∞ (2 x^2-3 x+5)(3 x-1)^ (x)/(2) (3 x^2+5 x+4) (3 x+2)^x...

\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 :

Options

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

Correct answer

C. $\frac{2}{3 \sqrt{\mathrm{e}}}$

Step-by-step 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}}

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