Problems from Indeterminate form infinity/infinity

Development:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{\frac{1}{2}x+56}{3x+1}}={\displaystyle \frac{\frac{1}{2}x}{3x}}={\displaystyle \frac{\frac{1}{2}}{3}}={\displaystyle \frac{1}{6}}}$

Solution:

${\displaystyle \frac{1}{6}}$

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Development:

a) ${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{-x^2+x+logx}{3x}}=\underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{-x^2}{3x}}=-(+\mathrm{\infty }{)}^2=-\mathrm{\infty }}$

b) ${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{x^3+3}{x}}=\underset{x\to -\mathrm{\infty }}{lim}{x}^3=(-\mathrm{\infty }{)}^3=-\mathrm{\infty }}$

Solution:

a) $-\mathrm{\infty }$

b) $-\mathrm{\infty }$

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