Question

Disque sacar el limte del infinito​

Answer 1

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim \: \frac{ {9x }^{3} + {2x}^{2} + 1 }{ {3x}^{5} - 5 } }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{ x→ \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim (\frac{ {x}^{3}(9 - \frac{2}{x} + \frac{1}{ {x}^{3} }) }{ {x}^{3}( 3 - \frac{5}{ {x}^{3} } ) }) }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{x → \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim (\frac{9 - \frac{2}{x} + \frac{1}{ {x}^{3} } }{3 - \frac{5}{ {x}^{3} } }) }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{x → \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{ \frac{9 - 2 \times 0 + 0}{3 - 5 \times 0} }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{ = 3}}{\hspace{0pt}\above 2.2pt}}$

________________________________

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim (\frac{ {x}^{2} + x - 1 }{ {2x}^{2} + 5 } )}}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{x → \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim (\frac{ {x}^{2} (1 + \frac{1}{x} - \frac{1}{ {x}^{2} } ) }{ {x}^{2}(2 + \frac{5}{ {x}^{2} } ) }) }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{x → \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{lim (\frac{1 + \frac{1}{x} - \frac{1}{ {x}^{2} } }{2 + \frac{5}{ {x}^{2} } }) }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{x → \infty }}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{ \frac{1 + 0 - 0}{2 + 5 \times 0} }}{\hspace{0pt}\above 2.2pt}} \\ \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{ = \frac{1}{2} }}{\hspace{0pt}\above 2.2pt}}$

_____________________________