anonymous
  • anonymous
CONTINUITY QUESTION : Examine the continuity of the function : f(x)= ( 2[x] )/ ( 3x-|x| ) at x=-1/2 and x=1
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
In order to get the best possible answers, it is helpful if you say in what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level.
anonymous
  • anonymous
Well said @FoolForMath ! :p
anonymous
  • anonymous
I know, I am so amazing ;)

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anonymous
  • anonymous
Oh please ! WAKE UP ! :P
anonymous
  • anonymous
Anyways , I will believe that you are SUPER-DUPER AMAZING , only when you will SOLVE THIS question ! :p And please no direct answers , the whole explanation is required :p
anonymous
  • anonymous
who said I need your credence? :P
anonymous
  • anonymous
A FRIEND IN NEED IS A FRIEND INDEED ! So no "faaltu" baatein, help ! :p
anonymous
  • anonymous
Sorry, I don't do baby sitting :P
anonymous
  • anonymous
Now from where on this earth "baby sitting" came ? Iam asking you to help me solve an ECO(H) , 2nd year maths question , not to sing a lullaby ! :p Had hai matlab ! ;p
anonymous
  • anonymous
Thanks for honouring me with your MEDAL ! :P
anonymous
  • anonymous
Anyways I gtg now , hope to see the answer posted ! :p Cya ! :-)
anonymous
  • anonymous
You should not worr y about the denominator. It is continious at any x except at x=0. All you have to examine is the function g(x) = [x]. Where is the function g(x) discontinuous?
anonymous
  • anonymous
See for example http://mathworld.wolfram.com/IntegerPart.html
anonymous
  • anonymous
check right and left limits at x=-1/2 and x=1 If right limits are equal to left limits the function is continuous
anonymous
  • anonymous
Iam not able to deal with [x] ! :-( When there is a positive number inside [x] , that's ok with me , but when there is a negative number , Iam not able to apply the left/right hand limit ! Please help
anonymous
  • anonymous
[-1.5] = -2. More formally, \[ \lfloor-a \rfloor = - \lceil a \rceil \]
anonymous
  • anonymous
You do not have negative numbers now, you have 1 and 1/2
anonymous
  • anonymous
its -1/2
anonymous
  • anonymous
Ok apply LHL to [x] where x->-1/2
anonymous
  • anonymous
I did not see the -. I need better glasses.
anonymous
  • anonymous
[ -1/2 ] = [-0.5]=-1
anonymous
  • anonymous
[x} is constant at any open interval having end points n, n+1 or n is a positive or negative integer.
anonymous
  • anonymous
so -1/2 is in the interval (-1,0). Is {x] continuous at -1/2?
anonymous
  • anonymous
Heya ! Iam finally through with this question. So before closing this question , I would like to thank all of you-" THANK YOU !" :-)

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