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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
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?
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 :)