anonymous
  • anonymous
Find the lim x->-infinity of (6-7x)/(3+2x)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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watchmath
  • watchmath
It is of the type \(\frac{\infty}{\infty}\) so we can use L'Hospital rule and reduce the limit into: \(\lim_{x\to -\infty}\frac{-7}{4(3+2x)^3}=0\)
watchmath
  • watchmath
sorry wrong derivative on the bottom :)
anonymous
  • anonymous
yeah i saw that but either way it doesnt matter.

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watchmath
  • watchmath
the bottom should be \(8(3+2x)^3\), but the limit is still 0.
myininaya
  • myininaya
\[\lim_{x \rightarrow -\inf}\frac{6-7x}{(3+2x)^4}\times \frac{x^4}{x^4}\]
myininaya
  • myininaya
top goes to 0 bottom gores to 2^4 so the limit is 0
myininaya
  • myininaya
i meant to write (1/x^4)/(1/x^4)
anonymous
  • anonymous
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.
myininaya
  • myininaya
yeah rob i didn't feel like expanding it lol
anonymous
  • anonymous
along should have been spelled alone. Sorry.

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