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

c0nman56

Answered question

2022-10-12

Behaviour of the function ln ( 1 + x 2 )
Thus function has derivative equal to:
2 x 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.

Answer & Explanation

cesantedz

cesantedz

Beginner2022-10-13Added 12 answers

You stumbled upon a classic counterexample to the "theorem"
f : R R has an horizontal asymptote if and only if f ( x ) 0 as x
Now, your function f ( x ) = log ( 1 + x 2 ) is a counterexample to the " " implication, since lim x f ( x ) = + . This can be shown by many means; but let's try with the definition, which is ε > 0 there exist a δ > 0 such that | log ( 1 + x 2 ) | > ε if x > δ
Now, log ( 1 + x 2 ) > ε means that x 2 > e ε 1; so we only have to take δ = e ε 1 to see that if x > δ = e ε 1 , then log ( 1 + x 2 ) > ε
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 y = f ( x 0 ) + f ( x 0 ) ( x x 0 ); if y = c is an horizontal asymptote for f, then it must be tangent to the graph of f ( x ) "at infinity". So now c = c + lim x 0 f ( x 0 ) ( x x 0 ), from which we conclude that lim x 0 f ( x 0 ) = 0
edit: of course for all this to be true we must have that lim x f ( x ) must exist, otherwise we can't prove anything as a comment points out.
Tara Mayer

Tara Mayer

Beginner2022-10-14Added 4 answers

Short answer: Logarithms don't have horizontal asymptotes.
Long answer: the derivative becomes smaller as x , but it has no horizontal asymptote. The asymptote you thought you found, is not an asymptote since ln ( 1 + x 2 ) = 14.25 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.

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