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anonymous

  • 5 years ago

Find the lim x->-infinity of (6-7x)/(3+2x)^4

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  1. watchmath
    • 5 years ago
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    It is of the type \(\frac{\infty}{\infty}\) so we can use L'Hospital rule and reduce the limit into: \(\lim_{x\to -\infty}\frac{-7}{4(3+2x)^3}=0\)

  2. watchmath
    • 5 years ago
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    sorry wrong derivative on the bottom :)

  3. anonymous
    • 5 years ago
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    yeah i saw that but either way it doesnt matter.

  4. watchmath
    • 5 years ago
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    the bottom should be \(8(3+2x)^3\), but the limit is still 0.

  5. myininaya
    • 5 years ago
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    \[\lim_{x \rightarrow -\inf}\frac{6-7x}{(3+2x)^4}\times \frac{x^4}{x^4}\]

  6. myininaya
    • 5 years ago
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    top goes to 0 bottom gores to 2^4 so the limit is 0

  7. myininaya
    • 5 years ago
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    i meant to write (1/x^4)/(1/x^4)

  8. anonymous
    • 5 years ago
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    The fraction with the denominator expanded. \[\frac{6-7x}{\left(81+216 x+216 x^2+96 x^3+16 x^4\right)} \] It appears that the 16x^4 term along will do the job.

  9. myininaya
    • 5 years ago
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    yeah rob i didn't feel like expanding it lol

  10. anonymous
    • 5 years ago
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    along should have been spelled alone. Sorry.

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