anonymous
  • anonymous
examine the continuity of the following function. f(x)=x-[x]. [x] represents the greatest integer less than or equal to x
Precalculus
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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abb0t
  • abb0t
the notation is: |dw:1370325042529:dw|
anonymous
  • anonymous
yeah.. it is actually like that but i dont have it. do you know the answer @abb0t
abb0t
  • abb0t
I don't even know your question. Lol

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abb0t
  • abb0t
But since you mentioned continuity, I;m guessing this is a limit's question?
anonymous
  • anonymous
yes. exactly.
anonymous
  • anonymous
it is about continuous and discontinuous functions
abb0t
  • abb0t
Well, x itself is continuous. Draw thegraph.
abb0t
  • abb0t
|dw:1370325386097:dw|
Valpey
  • Valpey
abb0t drew the ceiling function but what you described is the floor function. Most things with ceiling or floor functions are discontinuous.
anonymous
  • anonymous
@abb0t.. but the function has \[\lfloor x \rfloor\]
anonymous
  • anonymous
that's exactly my concern @Valpey.. how should i start?
Valpey
  • Valpey
The graph looks like: |dw:1370325561574:dw|
Valpey
  • Valpey
When f(x) jumps from 1 back down to 0 you have a discontinuity. A function is continuous if the limit as you approach any point from the left or the right is the same.
anonymous
  • anonymous
when we talk about limit exist, it has to be RHS=LHS, right?
abb0t
  • abb0t
they must be same as you approach from both sides (left and right)
Valpey
  • Valpey
Consider \[f(3+\epsilon)\] and \[f(3-\epsilon)\] where epsilon is a very small number. Show that as epsilon approaches zero, they don't converge.
anonymous
  • anonymous
then, since [x] represents as i mentioned above, so LHS will not be same as RHS
anonymous
  • anonymous
if we try to solve using limit? @Valpey
Valpey
  • Valpey
Basically, for epsilon positive:\[\lim_{\epsilon \rightarrow 0}f(3+\epsilon)= \lim_{\epsilon \rightarrow 0}(\epsilon) = 0\] Whereas \[\lim_{\epsilon \rightarrow 0}f(3 -\epsilon) = \lim_{\epsilon \rightarrow 0}(1-\epsilon) = 1\]
anonymous
  • anonymous
if we were to choose x=1 and limit approach to 1.. is it possible @Valpey?
anonymous
  • anonymous
if we choose x=1, then f(1)=0
ParthKohli
  • ParthKohli
@abb0t the notation is:|dw:1370423855576:dw|

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