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mathcalculus Group Title

HELP: evaluating limits to infinity by dividing x (ATTACHED BELOW)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. mathcalculus Group Title
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    i got 3/5 but it's wrong.

    • one year ago
  3. mathcalculus Group Title
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    @kropot72 do you know this?

    • one year ago
  4. mathcalculus Group Title
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    @satellite73 any ideas?

    • one year ago
  5. hihihii Group Title
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    0

    • one year ago
  6. hihihii Group Title
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    if the denominator grows bigger than the numerator, then the limit is 0

    • one year ago
  7. mathcalculus Group Title
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    oh so we don;t need to divide by x? or do anything to it?

    • one year ago
  8. hihihii Group Title
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    nope!

    • one year ago
  9. mathcalculus Group Title
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    thanks! :) @hihihii

    • one year ago
  10. hihihii Group Title
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    no problem!

    • one year ago
  11. mathstudent55 Group Title
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    \[\lim_{x \rightarrow \infty} \dfrac{3x + 2}{3x^2 - 5x + 8}\] \[ =\lim_{x \rightarrow \infty} \dfrac{ x^2 \left( \frac{3}{x} + \frac{2}{x^2} \right) } {x^2 \left( 3 - \frac{5}{x} + \frac{8}{x^2} \right) }\] \[ = \lim_{x \rightarrow \infty} \dfrac{ \frac{3}{x} + \frac{2}{x^2} } { 3 - \frac{5}{x} + \frac{8}{x^2} } \] \[ = \dfrac{ 0 + 0} { 3 - 0 + 0 } \] \(= 0\)

    • one year ago
  12. mathcalculus Group Title
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    thanks u @mathstudent55

    • one year ago
  13. mathstudent55 Group Title
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    You're welcome.

    • one year ago
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