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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.