Function with special behavior near zero

Question

I'm looking for functions that have the following behaviors:

$f(x) \to 0$ as $|x| \to 0$,
as $x \to 0^+$, $\alpha < \frac{{df(x)}}{{d(x)}}$ for any $0<\alpha<\infty $.

One example of this kind of functions is $f(x)=x^a$ where $0<a<1$.

Thank you for your help.

I assume (2) is $\lim\limits_{x\to0^+}f'(x)=+\infty$? – anon Jun 18 '15 at 02:49 — anon, Jun 18 '15 at 02:49
Yes, I expressed it in another way. – Reza H. Khayyami Jun 18 '15 at 02:52 — Reza H. Khayyami, Jun 18 '15 at 02:52

Indominus · Accepted Answer · 2015-06-18T03:28:42.557

2

1.

$f(x)=-x\ln{|x|}$. Verify:

$f(0^{+})=f(0^{-})=0$, so $\lim_{x\to0}f(x)=0$;

$f^{'}(x)=-\ln{|x|}-1$, so $\space f^{'}(0^{+})=+\infty$.

2.

$f(x)=\frac{\arctan{x}}{\sqrt{|x|}}$.Verify:

$f(0^{+})=f(0^{-})=0$ (L'Hôpital's rule: $\lim_{x\to0^{+}}f(x)=lim_{x\to0^{+}}\frac{\frac{1}{1+x^2}}{\frac{1}{2\sqrt{x}}}=im_{x\to0^{+}}\frac{2\sqrt{x}}{1+x^2}=0$)

$f^{'}(x)=\frac{1}{(1+x^2)\sqrt{x}}-\frac{\arctan{x}}{2x\sqrt{x}}\to\infty\space$ as $x\to0^{+}$

edited Jun 18 '15 at 03:28

answered Jun 18 '15 at 02:57

Indominus

1,166

Nice, thanks alot for your effort. The second one is much more interesting as it is bounded. – Reza H. Khayyami Jun 18 '15 at 03:35
1

Actually, being bounded is not such an amazing feature, because nothing outside of the neighborhood of $0$ matters here. You can always make the function like the one in Example 1 bounded, by "cutting its tails". – Indominus Jun 18 '15 at 03:43
Yes, you are right, however in some applications bounded functions is demanded. – Reza H. Khayyami Jun 18 '15 at 03:57
I just generalize your second suggestion as follows:
$f(x) = {\left| x \right|^a}\tan (x)$ where $-1<a<0$.
– Reza H. Khayyami Jun 18 '15 at 04:11
If you want bounded, it should be $\arctan{x}$ instead of $\tan{x}$. Otherwise $\tan{x}$ also works. – Indominus Jun 18 '15 at 05:00
Yes it was typo, $f(x) = {\left| x \right|^a}\arctan(x)$ – Reza H. Khayyami Jun 18 '15 at 05:02

Function with special behavior near zero

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