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transgalactic
Dec18-08, 11:26 AM
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
??

HallsofIvy
Dec18-08, 01:42 PM
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.

transgalactic
Dec18-08, 02:22 PM
thanks

statdad
Dec18-08, 04:58 PM
Halls, I think you have a typo here:
...but f(0)= 1 so sup (least upper bound) of the range of f is 1.

because of the \ln x term, f(0) is undefined. I believe you meant to
type


f(1) = 1

HallsofIvy
Dec18-08, 05:01 PM
Thanks statdad. I have edited so I can pretend I never made that mistake!

statdad
Dec18-08, 05:02 PM
Nothing to see here. Move along. :smile:

LuisVela
Jan21-09, 01:51 PM
As x goes to either infinity or negative infinity f(x) goes to 0.

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