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anonymous

  • 5 years ago

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

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  1. dumbcow
    • 5 years ago
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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

  2. anonymous
    • 5 years ago
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    what happened with -2x in 4th line?

  3. dumbcow
    • 5 years ago
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    oh i made the exponent positive by putting it in the denominator 1/x = x^-1

  4. anonymous
    • 5 years ago
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    Is this limes 0 because e^2x goes faster to infinity that 1-2x does?

  5. dumbcow
    • 5 years ago
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    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?

  6. anonymous
    • 5 years ago
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    When I asked what happened with -2x, I thought of 1-2x ( there's no 1-2x in 4th line, why did it disappeared? )

  7. dumbcow
    • 5 years ago
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    um when i went lim 1-2x = 1-infinity ?? i was substituting infinity in for x to evaluate lim

  8. anonymous
    • 5 years ago
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    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?

  9. dumbcow
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    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?

  11. nikvist
    • 5 years ago
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  12. dumbcow
    • 5 years ago
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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

  13. anonymous
    • 5 years ago
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    Thank you both :)

  14. amistre64
    • 5 years ago
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    tag!!

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