Babynini
  • Babynini
Limits!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Babynini
  • Babynini
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Babynini
  • Babynini
@freckles :)
Babynini
  • Babynini
So direct substitution didn't work. Then I tried ((t-4)^4)/((t-4)^3) Does that work? where would I go from there?

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More answers

unimatix
  • unimatix
You want to plug in values increasing close to 4 for t.
unimatix
  • unimatix
like 3.9, 4.1 then 3.99, 4.01 etc... until you see the pattern and there is your answer
Babynini
  • Babynini
approaching 4 yeah?
unimatix
  • unimatix
yes number approaching 4
Babynini
  • Babynini
Ok, let me put it into my calc, just a second :)
Babynini
  • Babynini
does 5.33 look right?
Babynini
  • Babynini
hrm, it's saying that's wrong.
unimatix
  • unimatix
5.333(repeating) is right believe.
unimatix
  • unimatix
Maybe try imputing 16/3
Babynini
  • Babynini
Oh yay! It accepted 16/3 Thanks so much :)
unimatix
  • unimatix
You're welcome :)
jim_thompson5910
  • jim_thompson5910
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, \[\Large \frac{t^4-256}{t^3-64} = \frac{(t - 4)(t+4)(t^2 + 16)}{(t-4)(t^2 + 4t + 16)}\] \[\Large \frac{t^4-256}{t^3-64} = \frac{\cancel{(t - 4)}(t+4)(t^2 + 16)}{\cancel{(t-4)}(t^2 + 4t + 16)}\] \[\Large \frac{t^4-256}{t^3-64} = \frac{(t+4)(t^2 + 16)}{t^2 + 4t + 16}\]
jim_thompson5910
  • jim_thompson5910
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)
Babynini
  • Babynini
ahh that makes sense!! Thanks so much. Yeah, I was trying to figure out how to do it algebraically.

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