Answer:
• As x approaches -∞, g(x) approaches -∞; and as x approaches ∞, g(x) approaches ∞.
Step-by-step explanation:
[tex]{ \tt{g(x) = \frac{ {x}^{2} + 5x}{3x} }} \\ [/tex]
If x approaches -∞;
[tex]{ \tt{g( {}^{ - } \infin}) = \frac{ {( {}^{ - } \infin)}^{2} + {}^{ - } \infin}{ {}^{ - } \infin} } \\ \\ { \boxed{ \tt{g( {}^{ - } \infin) = {}^{ - } \infin }}}[/tex]
If x approaches +∞;
[tex]{ \tt{g( {}^{ + } \infin) = \frac{ {( \infin)}^{2} + \infin}{ \infin} }} \\ \\ { \boxed{ \tt{ g( {}^{ + } \infin) = {}^{ + } \infin }}}[/tex]
Note: I'm taking my infinite value to be 1,000
