Problems from Indeterminate form 0 x infinity

Development:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{x+1}{3x^2}}\cdot {\displaystyle \frac{x^2}{x-1}}=0\cdot \mathrm{\infty }}$

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

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{x\cdot x^2}{3x^2\cdot x}}={\displaystyle \frac{1}{3}}}$

Solution:

${\displaystyle \frac{1}{3}}$

Hide solution and development

Development:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{x}}{x^2}}\cdot (x^3+1)=0\cdot \mathrm{\infty }}$

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

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{x^3\sqrt{x}}{x^2}}=\underset{x\to +\mathrm{\infty }}{lim}x\sqrt{x}=+\mathrm{\infty }}$

Solution:

$+\mathrm{\infty }$

Hide solution and development