Finding the Lim inf and sup of f(x)=1/(1+(ln x)^2) for Infinity and Beyond

[transgalactic]

i need to find the sup and the inf of

<br /> f(x)=1/(1+(ln x)^2 )<br /> <br />

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]

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]

thanks

 

[statdad]

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

<br /> f(1) = 1<br />

 

[HallsofIvy]

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

 

[statdad]

Nothing to see here. Move along.

 

[LuisVela]

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.