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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
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