f:[-1,1] -> R is defined as .......| 1/ln(x) , x not equal be 0 f(x)={ .......| 0, x=0 how do i show this is continious??? thankz
Last edited by dopi; Nov 14th 2006 at 12:33 PM. Reason: error
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Originally Posted by dopi f:[-1,1] -> R is defined as .......| 1/ln(x) , x not equal be 0 f(x)={ .......| 0, x=0 how do i show this is continious??? thankz But ln(x) is only defined on (0, infinity)! Your function's domain is too large. -Dan
Originally Posted by topsquark But ln(x) is only defined on (0, infinity)! Your function's domain is too large. -Dan no its not ...its defined on [-1, 1] isnt it?
Maybe you mean: Now it is continuous.
Originally Posted by Plato Maybe you mean: Now it is continuous. yea thats wat i meant...i forgot to put the modulus sign on...but how would i show that this is continious with f:[-1,1] thankz
Still no good you should have mentioned the open interval Anyway, the way you show continuity is to show that, Since, open some open interval except possibly at we have, Which is true if we consider the left-right limits, And by using L'Hopital's Rule.
Originally Posted by ThePerfectHacker Still no good you should have mentioned the open interval Anyway, the way you show continuity is to show that, Since, open some open interval except possibly at we have, Which is true if we consider the left-right limits, And by using L'Hopital's Rule. but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help
Originally Posted by dopi but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help TPH was referring to the fact that so is undefined at . -Dan
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