Finding the Limit of x as t Tends to Infinity

[nokia8650]

"x = (e^(2kt) - 1)/(4e^(2kt) - 2)"

How would I find the limit of this expression as t tends to infinity?

As t --> infinity, the two exponentials also tend to infinity. However, that was as far as I could go. It is clear by subbing large values of t in, that the limit should be 1/4, however I am unable to prove this.

Thanks

 

[Vid]

Factor e^(2kt) out of the top and bottom.

 

[HallsofIvy]

One of the first things you should have learned is to divide both numerator and denominator by the "largest" term- in this case e^(2kt) (that's what Vid was saying). You will be left with e^(-2kt) in each and that's easy.

You could also use L'Hopital's rule here but that is "overkill"

 

[Dick]

k is positive, I hope. Otherwise you'd better split into cases.

 

[Vid]

If k is negative, the only thing that changes is whether you add or subtract zero; there's no need for cases.

 

[Dick]

If k is negative, the only thing that changes is whether you add or subtract zero; there's no need for cases.

Add or subtract zero? There are three different cases. k>0, k=0 and k<0.

 

[Vid]

My statement about k being positive or negative is true,but I did forget about k=0, which does need its own case.

 

[nokia8650]

Thanks so much for the help. I forget to mention that k is positive. Therefore, I get:

x = (1-1/e(^2kt)) / ((4 - 2/e(^2kt))

Thus, as t tends to infinity, x will tend to 1/4.

However, say k were to be negative, how would it change? Would x tend to 1?

Thanks

 

[Dick]

If k is negative then exp(2kt)->0. Then you can just drop the exponentials and get 1/2.