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kpham

evaluate: lim x->2 sqrt (6-x)-2/sqrt (3-x)-1

  • one year ago
  • one year ago

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  1. kpham
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    valuate: lim x->2 [sqrt (6-x)]-2/[sqrt (3-x)]-1

    • one year ago
  2. kpham
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    help me plzz

    • one year ago
  3. zzr0ck3r
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    do u know lhopital rule?

    • one year ago
  4. kpham
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    whats that

    • one year ago
  5. kpham
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    im learning about limits and valuate limit

    • one year ago
  6. rainbow22
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    Square top and bottom?

    • one year ago
  7. kpham
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    top is sqrt of 6-x only bottom is sqrt of 3-x only

    • one year ago
  8. febylailani
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    two sided limit doesnt exist.

    • one year ago
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