lim(1-2x)e^(-2x), x->infinity

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lim(1-2x)e^(-2x), x->infinity

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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this can be written as lim (1-2x) * lim 1/e^2x lim 1-2x = 1-infinity = -infinity lim 1/e^2x = 1/e^infinity = 1/infinity = 0 -> -infinity*0 = 0 lim = 0
what happened with -2x in 4th line?
oh i made the exponent positive by putting it in the denominator 1/x = x^-1

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Is this limes 0 because e^2x goes faster to infinity that 1-2x does?
the 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 1-2x) it would still go to 0 it would just take longer make sense?
When I asked what happened with -2x, I thought of 1-2x ( there's no 1-2x in 4th line, why did it disappeared? )
um when i went lim 1-2x = 1-infinity ?? i was substituting infinity in for x to evaluate lim
ok, 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?
its 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
No. 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?
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i stand corrected, thank you ok so if we use L'Hospital's rule first we write it as lim (1-2x)/e^2x = -infinity/infinity ->indeterminate differentiate top and bottom lim -2/2e^2x = -2/infinity = 0 so lim is still 0
Thank you both :)
tag!!

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