Indeterminate form 0 x infinity

$$\displaystyle\lim_{x \to{+}\infty}{\dfrac{\sqrt{x}}{x^2}\cdot(x^3+1)}=0\cdot\infty$$

As this limit $x^3+1\approx x^3$

$$\displaystyle\lim_{x \to{+}\infty}{\dfrac{x^3\sqrt{x}}{x^2}}=\lim_{x \to{+}\infty}{x\sqrt{x}}=+\infty$$

$$\displaystyle\lim_{x \to{+}\infty}{\dfrac{x+1}{3x^2}\cdot\dfrac{x^2}{x-1}}=0\cdot\infty$$

As this limit $x+1\approx x$ y $x-1\approx x$

$$\displaystyle\lim_{x \to{+}\infty}{\dfrac{x\cdot x^2}{3x^2\cdot x}}=\dfrac{1}{3}$$