# Lim inf/sup question

1. Dec 18, 2008

### transgalactic

i need to find the sup and the inf of

$f(x)=1/(1+(ln x)^2 )$

i can find the limit for the function for + infinity
and - infinity

but what to do next?
the sup is the least upper bound
the limit for +infinity is not SUP
??

2. Dec 18, 2008

### HallsofIvy

Staff Emeritus
I assume you mean sup and inf of the range of that function. You seem to be thinking that the limit as x goes to infinity SHOULD be the sup and there is no reason for that. As x goes to either infinity or negative infinity f(x) goes to 0. Since f is never negative, 0 is the inf (greatest lower bound) of the range of f. Since the denominator is always larger than or equal to the numerator, f is never larger than 1- but f(1)= 1 so sup (least upper bound) of the range of f is 1.

Last edited: Dec 18, 2008
3. Dec 18, 2008

### transgalactic

thanks

4. Dec 18, 2008

Halls, I think you have a typo here:
because of the $$\ln x$$ term, $$f(0)$$ is undefined. I believe you meant to
type

$$f(1) = 1$$

5. Dec 18, 2008

### HallsofIvy

Staff Emeritus
Thanks statdad. I have edited so I can pretend I never made that mistake!

6. Dec 18, 2008

Nothing to see here. Move along.

7. Jan 21, 2009

### LuisVela

....ln(x) is not defined for x<=0.
The domain of f(x) is {x:x>0}.
Cheers.