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clara1223

  • one year ago

given that f(x)=(x^2)-4x, evaluate the limit as x approaches 2 of (f(x)-f(3))/(x-2)

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  1. anonymous
    • one year ago
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    first you have to input f(x) into the equation, can you do that?

  2. clara1223
    • one year ago
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    yes, I got f(3)=-1, and i input f(x) and f(3) into the equation and got ((x^2)-4x-3)/(x-2)

  3. clara1223
    • one year ago
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    f(3)=-3 not -1

  4. anonymous
    • one year ago
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    that is correct, now you must factor the top, can you do that?

  5. anonymous
    • one year ago
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    CORRECTION: the top portion is actually x^2 - 4x + 3 because you're subtracting f(3)

  6. amistre64
    • one year ago
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    are you sure its f(3) ?

  7. jdoe0001
    • one year ago
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    hmmm

  8. clara1223
    • one year ago
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    oops, didn't notice that it was minus f(3)

  9. clara1223
    • one year ago
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    so that leaves me with (x-1)(x-3)/(x-2)

  10. jdoe0001
    • one year ago
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    well... hmm f(3) is -3, -f(3) is +3

  11. clara1223
    • one year ago
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    how do I get x-2 out of the denominator? or do I leave it there and the answer is that the limit doesn't exist?

  12. jdoe0001
    • one year ago
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    but I gather that's what amistre64 meant, using f(3), doesn't give us much relief on cancelling x-2

  13. clara1223
    • one year ago
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    it's definitely f(3)

  14. anonymous
    • one year ago
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    It does not exist, if you check the graph you can see that approaching 2, one side goes towards -infinity, and one goes to +infinity

  15. anonymous
    • one year ago
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    How about the lim = infinity?

  16. clara1223
    • one year ago
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    That's what I thought, thanks for confirming everyone!

  17. amistre64
    • one year ago
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    some foreshadowing tho: \[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\] is something that will come up alot later on

  18. anonymous
    • one year ago
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    there is a vertical asymptote at x=2 and it is non-removable if you need an explanation

  19. anonymous
    • one year ago
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    @meagannlaurell the limit exists but at infinity, it is different from "does not exist"

  20. clara1223
    • one year ago
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    In my class we interchange infinity and does not exist

  21. anonymous
    • one year ago
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    Well looking at the graph, the two sides approach different infinities, one negative and one positive. So it would still be DNE, and yes, it is debatable as to whether DNE should equal -/+infinity, but many classes do view them as interchangeable

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