# How do you graph [ln(1+x^3)]/x ?

Jan 3, 2017

Graph is inserted.

#### Explanation:

To make ln real, ${x}^{3} + 1 > 0$, giving $x > - 1$. The graph has horizontal asymptote$y = 0 \rightarrow$ and vertical asymptote $y = - 1 \uparrow$.

$y = \ln \frac{1 + {x}^{3}}{x} > 0 , x \in \left(- 1 , \infty\right)$, sans x = 0.

Despite that y is indeterminate ( $\frac{0}{0}$ ) at x = 0 and is having hole

of infinitesimal void that cannot be depicted in the graph.

$y = \ln \frac{1 + {x}^{3}}{x} \to 0$, as $x \to {0}_{\pm}$.

As $x \to \infty , y \to 0$.

As $x \to - {1}_{+} , y \to \frac{- \infty}{- 1} = \infty$.
graph{ln(1+x^3)/x [-10, 10, -5, 5]}