$\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{x / 2}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} \
& =\lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{5}{x^2}\right)(3 x)^{\frac{x}{2}}}{x^2\left(3+\frac{5}{x}+\frac{4}{x^2}\right)(3 x)^{\frac{x}{2}}} \cdot \frac{\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}}{\left(1+\frac{2}{3 x}\right)^{\frac{x}{2}}} \
& \lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}=e^{\lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}-1\right) \times \frac{x}{2}}=e^{\frac{-1}{6}} \
& \lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}\right)=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}-1\right) \times \frac{x}{2}} \
& =e^{\frac{1}{3}}
\end{aligned}So,\frac{2}{3} \times \frac{e^{\frac{-1}{6}}}{e^{\frac{1}{3}}}=\frac{2}{3} \times \frac{1}{e^{\frac{1}{3}+\frac{1}{6}}}=\frac{2}{3 \sqrt{e}}$