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anonymous
 5 years ago
lim(12x)e^(2x), x>infinity
anonymous
 5 years ago
lim(12x)e^(2x), x>infinity

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this can be written as lim (12x) * lim 1/e^2x lim 12x = 1infinity = infinity lim 1/e^2x = 1/e^infinity = 1/infinity = 0 > infinity*0 = 0 lim = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what happened with 2x in 4th line?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i made the exponent positive by putting it in the denominator 1/x = x^1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this limes 0 because e^2x goes faster to infinity that 12x does?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the limit is 0 because one part of the product goes to 0 how fast it gets there depends on which goes to infinity faster like you said but even if it were not the case (lets just say e^2x went slower than 12x) it would still go to 0 it would just take longer make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When I asked what happened with 2x, I thought of 12x ( there's no 12x in 4th line, why did it disappeared? )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0um when i went lim 12x = 1infinity ?? i was substituting infinity in for x to evaluate lim

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I understand that, u make product... but how do u know that infinity*0=0? isn't it undefined expression? We shouldn't apply L'Hospital's rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its one of the properties of numbers any number multiplied by 0 equals 0 in any other case infinity*any other num is undefined L'Hospital's rule can only be used for an indeterminate number such as 0/0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. you are wrong. Undefined expressions are 0/0, ∞/∞, ∞  ∞, 0▪ ∞, 1∞, 0°, ∞º So I think we shoul use L'Hospital here. Can you check it please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i stand corrected, thank you ok so if we use L'Hospital's rule first we write it as lim (12x)/e^2x = infinity/infinity >indeterminate differentiate top and bottom lim 2/2e^2x = 2/infinity = 0 so lim is still 0
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