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anonymous
 5 years ago
Find the lim x>infinity of (67x)/(3+2x)^4
anonymous
 5 years ago
Find the lim x>infinity of (67x)/(3+2x)^4

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1It 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
 5 years ago
Best ResponseYou've already chosen the best response.1sorry wrong derivative on the bottom :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i saw that but either way it doesnt matter.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1the bottom should be \(8(3+2x)^3\), but the limit is still 0.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \inf}\frac{67x}{(3+2x)^4}\times \frac{x^4}{x^4}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0top goes to 0 bottom gores to 2^4 so the limit is 0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i meant to write (1/x^4)/(1/x^4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The fraction with the denominator expanded. \[\frac{67x}{\left(81+216 x+216 x^2+96 x^3+16 x^4\right)} \] It appears that the 16x^4 term along will do the job.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yeah rob i didn't feel like expanding it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0along should have been spelled alone. Sorry.
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