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Lim inf/sup question

  • #1
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i need to find the sup and the inf of

[itex]
f(x)=1/(1+(ln x)^2 )

[/itex]

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
??
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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 by a moderator:
  • #4
statdad
Homework Helper
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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 [tex] \ln x [/tex] term, [tex] f(0) [/tex] is undefined. I believe you meant to
type

[tex]
f(1) = 1
[/tex]
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Thanks statdad. I have edited so I can pretend I never made that mistake!
 
  • #6
statdad
Homework Helper
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Nothing to see here. Move along. :smile:
 
  • #7
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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.
 

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