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show that lim (ye^(kt))/(1-y+ye^(kt)) which t ---> infinity?

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  • Wal C
    Lv 6
    1 decade ago
    Favorite Answer

    lim t→∞ [ (ye^(kt))/(1 - y + ye^(kt))]

    = lim t→∞ [ (ye^(kt))/(1 - y + ye^(kt)) * e^(-kt)/e^(k-t) ]

    = lim t→∞ [ y/(e^(-kt) - ye^(-kt) +y)]

    = y/(0 - y*0 + y) (since, as t→∞, e^(-kt) →0)

    = y/y

    = 1

    Source(s): Me ;^))
  • 1 decade ago

    Limit as t tends to infinity of (ye^(kt))/(1-y+ye^(kt))

    Divide the numerator and each denominator term by ye^(kt).

    This gives:

    1 / [1/(ye^(kt)) - 1/e^(kt) + 1]

    As t tends to infinity, both ye^(kt) and e^(kt) tend to infinity.

    Therefore, their reciprocals will tend to zero.

    Thus, the expression becomes :

    1 / [0 - 0 + 1] = 1

  • 1 decade ago

    lim t--->infinity (ye^(kt))/(1-y+ye^(kt))

    For large values of t,the exponential term dominates so it becomes aproximately

    ye^kt/(ye^kt)=1

    for t---> infinity this approaches being exact

  • 1 decade ago

    After having calculated the derivatives for both terms of this fraction, you should realize you're searching for the limit of a constant number, that is 1.

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