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Compgroupmail

  • 3 years ago

Help- find the limit of:

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  1. Compgroupmail
    • 3 years ago
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    \[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ x^2+2x-8 } \right)+\left(\frac{ x-6 }{ x^2-4} \right)\]

  2. Compgroupmail
    • 3 years ago
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    so I found that \[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ (x+4)(x-2) } \right)+\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\] Now what?????

  3. nubeer
    • 3 years ago
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    now subsitute 2 in place of x

  4. Compgroupmail
    • 3 years ago
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    it will be 6/0. error. I can't do that yet.

  5. nubeer
    • 3 years ago
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    1/0 = infinity

  6. Compgroupmail
    • 3 years ago
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    sorry 1/3. I see the next step in the book but I don't understand how they got to it. I will write it up.

  7. Compgroupmail
    • 3 years ago
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    \[\lim_{x \rightarrow 2}\left( \frac{ 6(x+2)+(x-6)(x+4) }{ (x+4)(x-2)(x+2) } \right)\]

  8. nubeer
    • 3 years ago
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    they took the LCM in this step

  9. Compgroupmail
    • 3 years ago
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    ok. Can you explain that part to me? I don't seem to get it.

  10. nubeer
    • 3 years ago
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    write all the terms in denominator if some term is repeating more then once , just write it one time. like in this case on one side u have (x+4)(x−2) and on other side (x+2)(x−2) so write them all in denominator but (x-2) is written twice so u will just write it one time

  11. Compgroupmail
    • 3 years ago
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    ok but what about the numerator? How did they get to that part?

  12. nubeer
    • 3 years ago
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    ok now move back a step and see see what u had in denominator before . u had (x+4)(x−2) right?

  13. nubeer
    • 3 years ago
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    so you compare the old denominator with new numerator whichever is the common denominator u cancel them and whatever is left u multiply it with numerator

  14. nubeer
    • 3 years ago
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    |dw:1353067137453:dw|

  15. nubeer
    • 3 years ago
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    |dw:1353067229340:dw|

  16. nubeer
    • 3 years ago
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    |dw:1353067272848:dw|

  17. nubeer
    • 3 years ago
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    |dw:1353067293846:dw|

  18. nubeer
    • 3 years ago
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    are you getting what i am doing?

  19. Compgroupmail
    • 3 years ago
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    yes, so basically you devide the entire equation the the common denoiminator and then solve from there. I think I got it. Let me see if it works out.

  20. Zarkon
    • 3 years ago
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    \[\lim_{x \rightarrow 2} \left[\left( \frac{ 6 }{ (x+4)(x-2) } \right)+\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\right]\] \[\lim_{x \rightarrow 2} \left[\frac{x+2}{x+2}\left( \frac{ 6 }{ (x+4)(x-2) } \right)+\frac{x+4}{x+4}\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\right]\]

  21. Compgroupmail
    • 3 years ago
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    thnk

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