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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
Source(s): Me ;^)) - 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
- 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.