Question

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

Answer 1

The function has no asymptotes.

Explanation:

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]}