How do you graph $\frac{ln (1 + x^{3})}{x}$ $[ln(1+x^3)]/x$ ?

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How do you graph $\frac{ln (1 + x^{3})}{x}$ $[ln(1+x^3)]/x$ ?

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Answer 1 · 2017-01-03T14:44:35

Graph is inserted.

Explanation:

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

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

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

of infinitesimal void that cannot be depicted in the graph.

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

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

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

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How do you graph $\frac{ln (1 + x^{3})}{x}$ $[ln(1+x^3)]/x$ ?

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