c0nman56

2022-10-12

Behaviour of the function $\mathrm{ln}(1+{x}^{2})$

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.

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

Beginner2022-10-13Added 12 answers

You stumbled upon a classic counterexample to the "theorem"

$f:\mathbb{R}\to \mathbb{R}$ has an horizontal asymptote if and only if ${f}^{\prime}(x)\to 0$ as $x\to \mathrm{\infty}$

Now, your function $f(x)=\mathrm{log}(1+{x}^{2})$ is a counterexample to the "$\Leftarrow $" implication, since $\underset{x\to \mathrm{\infty}}{lim}f(x)=+\mathrm{\infty}$. 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}(1+{x}^{2})|>\epsilon $ if $x>\delta $

Now, $\mathrm{log}(1+{x}^{2})>\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}(1+{x}^{2})>\epsilon $

What about the other direction, "$\Rightarrow $"? 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 $y=f({x}_{0})+{f}^{\prime}({x}_{0})(x-{x}_{0})$; if $y=c$ is an horizontal asymptote for $f$, then it must be tangent to the graph of $f(x)$ "at infinity". So now $c=c+\underset{{x}_{0}\to \mathrm{\infty}}{lim}{f}^{\prime}({x}_{0})(x-{x}_{0})$, from which we conclude that $\underset{{x}_{0}\to \mathrm{\infty}}{lim}{f}^{\prime}({x}_{0})=0$

edit: of course for all this to be true we must have that $\underset{x\to \mathrm{\infty}}{lim}{f}^{\prime}(x)$ must exist, otherwise we can't prove anything as a comment points out.

$f:\mathbb{R}\to \mathbb{R}$ has an horizontal asymptote if and only if ${f}^{\prime}(x)\to 0$ as $x\to \mathrm{\infty}$

Now, your function $f(x)=\mathrm{log}(1+{x}^{2})$ is a counterexample to the "$\Leftarrow $" implication, since $\underset{x\to \mathrm{\infty}}{lim}f(x)=+\mathrm{\infty}$. 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}(1+{x}^{2})|>\epsilon $ if $x>\delta $

Now, $\mathrm{log}(1+{x}^{2})>\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}(1+{x}^{2})>\epsilon $

What about the other direction, "$\Rightarrow $"? 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 $y=f({x}_{0})+{f}^{\prime}({x}_{0})(x-{x}_{0})$; if $y=c$ is an horizontal asymptote for $f$, then it must be tangent to the graph of $f(x)$ "at infinity". So now $c=c+\underset{{x}_{0}\to \mathrm{\infty}}{lim}{f}^{\prime}({x}_{0})(x-{x}_{0})$, from which we conclude that $\underset{{x}_{0}\to \mathrm{\infty}}{lim}{f}^{\prime}({x}_{0})=0$

edit: of course for all this to be true we must have that $\underset{x\to \mathrm{\infty}}{lim}{f}^{\prime}(x)$ 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 $x\to \mathrm{\infty}$, but it has no horizontal asymptote. The asymptote you thought you found, is not an asymptote since $\mathrm{ln}(1+{x}^{2})=14.25$ 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.

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 $\mathrm{ln}(1+{x}^{2})=14.25$ 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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