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Babynini

  • one year ago

Limits!

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @freckles :)

  3. Babynini
    • one year ago
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    So direct substitution didn't work. Then I tried ((t-4)^4)/((t-4)^3) Does that work? where would I go from there?

  4. unimatix
    • one year ago
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    You want to plug in values increasing close to 4 for t.

  5. unimatix
    • one year ago
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    like 3.9, 4.1 then 3.99, 4.01 etc... until you see the pattern and there is your answer

  6. Babynini
    • one year ago
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    approaching 4 yeah?

  7. unimatix
    • one year ago
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    yes number approaching 4

  8. Babynini
    • one year ago
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    Ok, let me put it into my calc, just a second :)

  9. Babynini
    • one year ago
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    does 5.33 look right?

  10. Babynini
    • one year ago
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    hrm, it's saying that's wrong.

  11. unimatix
    • one year ago
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    5.333(repeating) is right believe.

  12. unimatix
    • one year ago
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    Maybe try imputing 16/3

  13. Babynini
    • one year ago
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    Oh yay! It accepted 16/3 Thanks so much :)

  14. unimatix
    • one year ago
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    You're welcome :)

  15. jim_thompson5910
    • one year ago
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    t^4 - 256 = (t^2)^2 - (16)^2 t^4 - 256 = (t^2 - 16)(t^2 + 16) t^4 - 256 = (t - 4)(t+4)(t^2 + 16) t^3 - 64 = t^3 - 4^3 t^3 - 64 = (t-4)(t^2 + 4t + 16) ... difference of cubes rule So, \[\Large \frac{t^4-256}{t^3-64} = \frac{(t - 4)(t+4)(t^2 + 16)}{(t-4)(t^2 + 4t + 16)}\] \[\Large \frac{t^4-256}{t^3-64} = \frac{\cancel{(t - 4)}(t+4)(t^2 + 16)}{\cancel{(t-4)}(t^2 + 4t + 16)}\] \[\Large \frac{t^4-256}{t^3-64} = \frac{(t+4)(t^2 + 16)}{t^2 + 4t + 16}\]

  16. jim_thompson5910
    • one year ago
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    after that, you can plug in t = 4 because there are no division by zero errors any more (the t-4 term in the denominator is gone)

  17. Babynini
    • one year ago
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    ahh that makes sense!! Thanks so much. Yeah, I was trying to figure out how to do it algebraically.

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