JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 12)
$\lim \limits_{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}$ is equal to :
$\frac{2 e}{3}$
$\frac{2}{3 \sqrt{\mathrm{e}}}$
$\frac{2 \mathrm{e}}{\sqrt{3}}$
$\frac{2}{\sqrt{3 \mathrm{e}}}$
Explanation
$$\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right)\left(1-\frac{1}{3 x}\right)^{x / 2}}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right)\left(1+\frac{2}{3 x}\right)^{x / 2}} \\
& =\lim _{x \rightarrow \infty} \frac{2}{3} \cdot \frac{e^{\frac{x}{2}\left(1-\frac{1}{3 x}-1\right)}}{e^{\frac{x}{2}\left(1+\frac{2}{3 x}-1\right)}} \\
& =\frac{2}{3} \cdot \frac{e^{-\frac{1}{6}}}{e^{1 / 3}}=\frac{2}{3} e^{-\frac{1}{2}}
\end{aligned}$$
Comments (0)
