Looking for a function

katkat

Does anyone know a function t(x) satisfying the following conditions.

as x -> 0 we have

(1) t(x)-> oo

(2) x t(x) -> 0

(3) x t(x) exp(t(x)) -> 0

(4) x t(x)^2 exp(t(x)) -> oo

I have "two extremes" as

t(x) = sqrt(-log x) satisfies (1-3) but x t(x)^2 exp(t(x)) -> 0

t(x) = -log( -x log(x))

satisfies (1-2) and (4), but x t(x) exp(t(x)) -> c >0

I have been searching desperately.

whozum

What level course is this?

steven187

hello

if your really desperate you should try ploting it through mathematica or some similar program and watch what happens as x goes to infinity it shouldnt be too hard, a few guesses and you should be able to get your solution

rachmaninoff

This is in the reals, right? And am I misreading this:
as x -> 0 we have

(1) t(x)-> oo

(2) x t(x) -> 0
or are (1) and (2) mutually exclusive?

whozum

rachmaninoff said:
This is in the reals, right? And am I misreading this:

or are (1) and (2) mutually exclusive?
I want to agree with you but first I want to verify that the only functions that tend to infinity as x-> 0 are ones where the denominator has a quicker tendency to infinity than the numerator.

joeboo

rachmaninoff said:
This is in the reals, right? And am I misreading this:

or are (1) and (2) mutually exclusive?

$$t(x) = \frac{1}{\sqrt{\vert x \vert}}$$

Am I missing something?

whozum

$$\lim_{x\rightarrow 0} \frac{x}{\sqrt{|x|}} \neq 0$$

rachmaninoff

edit: I misread the OP, ignore this post.

Last edited by a moderator:

shmoe

Homework Helper
whozum said:
$$\lim_{x\rightarrow 0} \frac{x}{\sqrt{|x|}} \neq 0$$
It's zero. When x>0 it's sqrt(x). It appears he's only looking at right handed limts (hence his log's), but for this function the two-sided limit is also zero, it's -sqrt(x) for x<0.

These limits are as x->0, none of them exclude any others.

t(x) = -log( -x log(x)) is hopeful. In condition 3, you're looking at [-log(x)-log(-log(x))]/(-log(x)). You'd like to speed up the growth of the denominator without affecting the numerator signifigantly. Try fiddling with the inside log of t(x)

whozum

I thought of this last night, does it have to be continuous?
And also, for a limit to exist, the function must approach the value from both sides, however since the log function can't take negative values, there is a whole in his domain for t(x) = -log( -x log(x)), correct?

shmoe

Homework Helper
The log(x) is what made me think he was considering right hand limits. Finding a function satisfying those limits from the right is enough in any case, just put |x| where you see x and you've got your two sided limits as well. There's nothing about continuity as he's written it, just a limit. In fact the first condition rules on continuity at 0. The t(x) = -log( -x log(x)) will work, with some minor alterations.

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