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anonymous

  • 5 years ago

limits part 2

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  1. anonymous
    • 5 years ago
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    graph attached please have a look

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  2. anonymous
    • 5 years ago
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    on the top is the question below is similar example need 4 answers

  3. watchmath
    • 5 years ago
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    Is the handwritten argument your answer? That's right!! :)

  4. anonymous
    • 5 years ago
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    1) f(2) exists (infact f(2) = 4)

  5. anonymous
    • 5 years ago
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    haha thats teachers the one on top is the hw quiz

  6. watchmath
    • 5 years ago
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    :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.

  7. anonymous
    • 5 years ago
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    2) lim f(x) does not exist as x -> 2 lim f(x) = DNE is that right?

  8. watchmath
    • 5 years ago
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    well the open circle and the fulls circle is above x=1. So you were not looking at the right point.

  9. anonymous
    • 5 years ago
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    its asking for 2 though that's why I wrote that

  10. watchmath
    • 5 years ago
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    yes, so don't look at the x=1.

  11. watchmath
    • 5 years ago
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    hold on the graph for your question is the one on the top right?

  12. anonymous
    • 5 years ago
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    yep

  13. watchmath
    • 5 years ago
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    Ok, so at x=2 we don't have any problem. There is no hole or jump at x=2. So it is continuous.

  14. anonymous
    • 5 years ago
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    right.

  15. anonymous
    • 5 years ago
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    part 2, limit exists from the right but not from the left

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