How do you find the asymptotes for $ln(x^2 + 1)$ ?

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Answer 1 · 2016-05-15T15:36:30

The function has no asymptotes.

The asymptote of the function $ln(x)$ occurs when $x=0$ .

So, for $ln(x^2+1)$ , this translates into asymptotes occurring whenever $x^2+1=0$ .

Attempting to solve $x^2+1=0$ gives $x^2=-1$ , which has no real solutions.

Thus, the graph of $ln(x^2+1)$ has no asymptotes. This is verified by a graph of $ln(x^2+1)$ :

graph{ln(x^2+1) [-31.48, 33.46, -11.76, 20.73]}

How do you find the asymptotes for ln(x^2 + 1)ln(x^2 + 1)?