Calculus 4

$\displaystyle\lim_{x \to \infty} \dfrac{2\sqrt{x}+2\sqrt[3]{x}+5\sqrt[5]{x}}{\sqrt{3x-2}+\sqrt[3]{2x-3}} \\ \quad \\ \displaystyle\lim_{x \to \infty} \dfrac{2+\dfrac{2}{x^{\frac16}}+\dfrac{5}{x^{\frac3{10}} }}{\sqrt{3-\dfrac{2}{x}}+\sqrt[3]{\dfrac{2}{x^{\frac12}}-\dfrac{3}{x^{\frac32}} }} \implies \dfrac{2}{\sqrt3} \implies \boxed{1.154}$

In the numerator we must neglect $3\sqrt[3]{x}+\sqrt[5]{x}$ as $2\sqrt x$ is much much greater compared to it as $x$ approaches $\infty.$ In the denominator,we must neglect $\sqrt[3]{2x-3}$ as $\sqrt {3x-2}$ is much much greater compared to it as $x$ approaches $\infty.$ Now the problem reduces to, $\lim_{x\to \infty} \frac{2\sqrt x}{\sqrt {3x-2}}.$ We again can neglect $-2$ in the denominator as much much smaller compared to $3x.$
So, $\lim_{x\to \infty}\frac{2\sqrt x}{\sqrt{3x}}$ $=\frac{2}{\sqrt3}=\boxed {1.154}.$