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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.

Git Gud
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Saurabh Raje
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    Why flatten out while approaching infinity would imply there is an asymptote? Take even $\ln x$ then. And it's obvious that, as $x\rightarrow \infty$, $\ln(1+x^2) \sim 2\ln x$. – Jean-Claude Arbaut Aug 24 '14 at 09:08

2 Answers2

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You stumbled upon a classic counterexample to the "theorem"

$f\colon \mathbb{R} \to \mathbb{R}$ has an horizontal asymptote if and only if $f'(x) \to 0$ as $x\to \infty$.

Now, your function $f(x) = \log(1+x^2)$ is a counterexample to the "$\Leftarrow$" implication, since $\lim_{x\to \infty} f(x) = + \infty$. This can be shown by many means; but let's try with the definition, which is $\forall \varepsilon > 0$ there exist a $\delta > 0$ such that $\left| \log(1+x^2) \right| > \varepsilon$ if $x > \delta$.

Now, $\log(1+x^2) > \varepsilon$ means that $x^2 > e^\varepsilon -1$; so we only have to take $\delta = \sqrt{e^\varepsilon -1}$ to see that if $x > \delta = \sqrt{e^\varepsilon -1}$, then $\log(1+x^2)> \varepsilon$.

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 $y = f(x_0) + f'(x_0)(x-x_0)$; if $y = c$ is an horizontal asymptote for $f$, then it must be tangent to the graph of $f(x)$ "at infinity". So now $c = c + \lim_{x_0 \to \infty} f'(x_0)(x-x_0)$, from which we conclude that $\lim_{x_0 \to \infty} f'(x_0) = 0$.

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

Andy
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  • I got the proof of the forward direction, but since the theorem is iff, shouldn't the backward direction be true as well? Why did I get a counter example for it? – Saurabh Raje Aug 24 '14 at 10:51
  • It is not an iff, since you found a counterexample! The point is that you might think it's a true statement (that's why I said "theorem" and not theorem), but it actually isn't. Another classical counterexample would be $f(x) = \sqrt{x}$, since $f'(x) = \frac{1}{2\sqrt{x}}$ tends to $0$ at infinity, but the square root function doesn't have an asymptote. – Andy Aug 24 '14 at 10:57
  • Both directions are false. For $\implies$ see this question. – Git Gud Aug 24 '14 at 11:43
  • @GitGud, you're absolutely right. The limit of the derivative must exist. I'll edit that in. – Andy Aug 24 '14 at 12:11
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Short answer: Logarithms don't have horizontal asymptotes.

Long answer: the derivative becomes smaller as $x\to\infty$, 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.

rae306
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