$\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{x / 2}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} \
& =\lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{5}{x^2}\right)(3 x)^{\frac{x}{2}}}{x^2\left(3+\frac{5}{x}+\frac{4}{x^2}\right)(3 x)^{\frac{x}{2}}} \cdot \frac{\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}}{\left(1+\frac{2}{3 x}\right)^{\frac{x}{2}}} \
& \lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}=e^{\lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}-1\right) \times \frac{x}{2}}=e^{\frac{-1}{6}} \
& \lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}\right)=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}-1\right) \times \frac{x}{2}} \
& =e^{\frac{1}{3}}
\end{aligned}So,\frac{2}{3} \times \frac{e^{\frac{-1}{6}}}{e^{\frac{1}{3}}}=\frac{2}{3} \times \frac{1}{e^{\frac{1}{3}+\frac{1}{6}}}=\frac{2}{3 \sqrt{e}}$
lim _x → ∞ (2 x^2-3 x+5 )(3 x-1)^ x 2 (3 x^2+5 x+4 ) √(3 x+2)^x is equal to :
JEE Main 2025 — Mathematics Calculus
x→∞lim(3x2+5x+4)(3x+2)x(2x2−3x+5)(3x−1)2x is equal to :
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
3e2
- B
32e
- C
3e2
- D
32e
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