Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

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

### 4 Answers

Relevance
• 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 ;^))
• 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

• 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

• 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.

Still have questions? Get your answers by asking now.