Mathematics - Limits Question with Solution | TestHub

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