c0nman56

2022-10-12

Behaviour of the function $\mathrm{ln}\left(1+{x}^{2}\right)$
Thus function has derivative equal to:
$\frac{2x}{1+{x}^{2}}$. 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.

cesantedz

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

Tara Mayer

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