anonymous
  • anonymous
limits part 2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
graph attached please have a look
1 Attachment
anonymous
  • anonymous
on the top is the question below is similar example need 4 answers
watchmath
  • watchmath
Is the handwritten argument your answer? That's right!! :)

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anonymous
  • anonymous
1) f(2) exists (infact f(2) = 4)
anonymous
  • anonymous
haha thats teachers the one on top is the hw quiz
watchmath
  • watchmath
:D Basically you need to compute three things similar to your earlier exercise. Compute the limit from left, right and the value of the function. Is one of them is different then the function is not continuous there.
anonymous
  • anonymous
2) lim f(x) does not exist as x -> 2 lim f(x) = DNE is that right?
watchmath
  • watchmath
well the open circle and the fulls circle is above x=1. So you were not looking at the right point.
anonymous
  • anonymous
its asking for 2 though that's why I wrote that
watchmath
  • watchmath
yes, so don't look at the x=1.
watchmath
  • watchmath
hold on the graph for your question is the one on the top right?
anonymous
  • anonymous
yep
watchmath
  • watchmath
Ok, so at x=2 we don't have any problem. There is no hole or jump at x=2. So it is continuous.
anonymous
  • anonymous
right.
anonymous
  • anonymous
part 2, limit exists from the right but not from the left

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