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cassieforlife5

  • one year ago

find the limit as f(g(x)) approaches 1 f(x)= 3/(x-1) g(x)= x^4 I got f(g(x))= 3/(x^4-1) but I need help finding the limit.

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  1. anonymous
    • one year ago
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    There's a vertical asymptote at 1. Have you tried taking the limits from the left and right sides?

  2. cassieforlife5
    • one year ago
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    on my graph I don't see a left hand limit, but is the right hand limit infinity?

  3. anonymous
    • one year ago
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    Yes the right is infinity Does your graph look like this?|dw:1442790635351:dw|

  4. cassieforlife5
    • one year ago
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    yes wait then from the left isn't the limit negative infinity

  5. anonymous
    • one year ago
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    right

  6. anonymous
    • one year ago
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    since the left and right limits aren't the same, the limit does not exist

  7. cassieforlife5
    • one year ago
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    ohhhh i see. Thanks so much! If you have time, would you mind helping me with another limit question? I'm not very good at limits

  8. anonymous
    • one year ago
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    sure

  9. cassieforlife5
    • one year ago
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    find the limit as x approaches 0 of (3x^4 - 6x^3)/(4x^3 + 2x^2)

  10. anonymous
    • one year ago
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    I'd start this one by factoring to see what cancels out.

  11. cassieforlife5
    • one year ago
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    okay so I got 3x^3(x-2) / 2x^2(2x+1)

  12. anonymous
    • one year ago
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    right, and two of the x's in the numerator will cancel the x² in the denominator, and then you can plug in the 0

  13. anonymous
    • one year ago
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    \[\frac{3x(x-2)}{2(2x+1)}\]

  14. cassieforlife5
    • one year ago
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    I got 0 as my answer

  15. anonymous
    • one year ago
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    that's right :)

  16. cassieforlife5
    • one year ago
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    Thanks a million! You're really a lifesaver!! :D

  17. anonymous
    • one year ago
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    you're welcome

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