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

### 4 Answers

- Wal CLv 61 decade agoFavorite 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

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- falzoonLv 71 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

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- 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

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- 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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