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anonymous

  • 5 years ago

evaluate limit as x goes to 2 = sin(x-2)/(x^2-4)

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  1. anonymous
    • 5 years ago
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    infinity

  2. anonymous
    • 5 years ago
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    take the limit of top and bottom sin(0) = 1 (2^2 - 4) aproaches 0 BUT can never go to zero because you cant divide by zero

  3. anonymous
    • 5 years ago
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    so as the denominator goes to 0.00000001 or 0.000000000000000001

  4. anonymous
    • 5 years ago
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    its ilike this 1/4 then 1/2 then 1/1 then 1/0.000001 then 1/0.0000000000001

  5. anonymous
    • 5 years ago
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    This one can be handled with a little manipulation. The bottom can be expanded to (x-2)(x+2). The lim of [sin (x-2)]/(x-2)]=1. That leaves 1/(x+2). So lim is 1/4.

  6. anonymous
    • 5 years ago
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    no the bottom cannot be expanded like that

  7. anonymous
    • 5 years ago
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    I am guessing by the way these questions are set up that x^2-4 is not sin of, but just a number.

  8. anonymous
    • 5 years ago
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    I think the its sin of x-2 so sin(x-2)

  9. anonymous
    • 5 years ago
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    divided by x^2 - 4

  10. anonymous
    • 5 years ago
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    In that case we are saying the same thing. \[x ^{2}-4=(x-2)(x+2)\]The difference of two squares.

  11. anonymous
    • 5 years ago
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    no thats what you are saying

  12. anonymous
    • 5 years ago
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    that is impossible to do

  13. anonymous
    • 5 years ago
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    I have had a couple beers. But are you telling me that the above is not the difference of two squares?

  14. anonymous
    • 5 years ago
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    no it is

  15. anonymous
    • 5 years ago
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    maybe your right im sorry

  16. anonymous
    • 5 years ago
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    limit as x goes to 2 = sin(x-2)/(x^2-4) 0/0 so Use L'Hopital's rule lim x->2 Cos(x-2)/(2x) = Cos (0) / (2*2) = 1/4

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