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

1 Answer
May 15, 2016

Answer:

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]}