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

Question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-03T14:44:35

Graph is inserted.

To make ln real, #x^3+1> 0#, giving #x > -1#. The graph has horizontal asymptote# y = 0 rarr# and vertical asymptote #y = -1 uarr#.

#y = ln(1+x^3)/x > 0, x in (-1, oo)#, sans x = 0.

Despite that y is indeterminate ( #0/0# ) at x = 0 and is having hole

of infinitesimal void that cannot be depicted in the graph.

#y = ln(1+x^3)/x to 0#, as #x to 0_(+-)#.

As # x to oo, y to 0#.

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