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

May 15, 2016

The function has no asymptotes.

#### Explanation:

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

So, for $\ln \left({x}^{2} + 1\right)$, 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 $\ln \left({x}^{2} + 1\right)$ has no asymptotes. This is verified by a graph of $\ln \left({x}^{2} + 1\right)$:

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