Behaviour of the function ln(1+x^2) Thus function has derivative equal to: (2x)/(1+x^2). This indicates that it will flatten out while approaching infinity, ie, should have an asymptote.Yet, the function does not have any real limiting value at infinity. To imvestigate further, I plotted the graph, which showed the asymptote/limiting value as 14.25 (you may check youtself if this is right) What can the reason be for not getting a real limit from the function itself? Please explain keeping in mind that I am in the last year of high school.
c0nman56
Answered question
2022-10-12
Behaviour of the function Thus function has derivative equal to: . This indicates that it will flatten out while approaching infinity, ie, should have an asymptote. Yet, the function does not have any real limiting value at infinity. To imvestigate further, I plotted the graph, which showed the asymptote/limiting value as 14.25 (you may check youtself if this is right) What can the reason be for not getting a real limit from the function itself? Please explain keeping in mind that I am in the last year of high school.
Answer & Explanation
cesantedz
Beginner2022-10-13Added 12 answers
You stumbled upon a classic counterexample to the "theorem" has an horizontal asymptote if and only if as Now, your function is a counterexample to the "" implication, since . This can be shown by many means; but let's try with the definition, which is there exist a such that if Now, means that ; so we only have to take to see that if , then What about the other direction, ""? Well, that is an actual theorem and we can prove it: to have an horizontal asymptote means to have a finite limit at infinity. Let's now remember that the tangent to the graph of a function is the line ; if is an horizontal asymptote for , then it must be tangent to the graph of "at infinity". So now , from which we conclude that edit: of course for all this to be true we must have that must exist, otherwise we can't prove anything as a comment points out.
Tara Mayer
Beginner2022-10-14Added 4 answers
Short answer: Logarithms don't have horizontal asymptotes. Long answer: the derivative becomes smaller as , but it has no horizontal asymptote. The asymptote you thought you found, is not an asymptote since has a solution (check it yourself). A way to think of it is: logarithms and exponential functions are inverses of each other. If logarithms had had a horizontal asymptote, then exponential functions would have had a vertical asymptote, which they don't have.