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anonymous

  • 5 years ago

Evaluate limit (x+1)/(2x-6)/(x^2-4) as x approaches 2

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  1. nowhereman
    • 5 years ago
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    You have to place more parentheses, to give that term a meaning!

  2. anonymous
    • 5 years ago
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    sorry

  3. anonymous
    • 5 years ago
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    its 3 things dividing each other

  4. anonymous
    • 5 years ago
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    so on top is \[((x+1)\div(2x-6))\div(x^2-4)\]

  5. anonymous
    • 5 years ago
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    is that better?

  6. anonymous
    • 5 years ago
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    its an improper fraction.

  7. nowhereman
    • 5 years ago
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    Yes. The limit does not exist, there is a pole at x=2.

  8. anonymous
    • 5 years ago
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    Right, there should be a removable discontinuity because the answer is -1/8

  9. nowhereman
    • 5 years ago
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    No, \(x^2 - 4\) tends to 0 and as it is in the denominator, plus the numerator is continuous at x=2, there must be a pole.

  10. anonymous
    • 5 years ago
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    So, the answer does not exist? Weird, because the answer key says that its -1/8.

  11. nowhereman
    • 5 years ago
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    Then the formula must be different.

  12. anonymous
    • 5 years ago
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    I simplified it a bit.

  13. anonymous
    • 5 years ago
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    The original formula is...

  14. anonymous
    • 5 years ago
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    \[((x/2)+(1/(x-3))/(x^2-4)\]

  15. nowhereman
    • 5 years ago
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    Then you must have made a mistake, because here the numerator tends to 0 too.

  16. anonymous
    • 5 years ago
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    I combined the numerator to find a common denominator for them.

  17. anonymous
    • 5 years ago
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    That's the original expression.

  18. anonymous
    • 5 years ago
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    I'm assuming I have to manipulate it to cancel out x-2 or x+2.

  19. anonymous
    • 5 years ago
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    and remove the discontinuity

  20. nowhereman
    • 5 years ago
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    Yes, \(x-2\). But it should work: \[\frac x 2 + \frac 1 {x-3} = \frac{x(x-3) + 2}{2(x-3)} = \frac{x^2-3x+2}{2(x-3)} = \frac{(x-2)(x-1)}{2(x-3)}\]

  21. anonymous
    • 5 years ago
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    perfect!

  22. anonymous
    • 5 years ago
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    Thank you.

  23. anonymous
    • 5 years ago
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    wait...

  24. anonymous
    • 5 years ago
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    nevermind.

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