• Welcome! The new FreeMathHelp.com forum is live. We've moved from VB4 to Xenforo 2.1 as our underlying software. Hopefully you find the upgrade to be a positive change. Please feel free to reach out as issues arise -- things will be a little different, and minor issues will no doubt crop up.

Can't understand this concept: lim_{x->infty} P(x)=(KP_0 e^{2x})/(K+P_0(e^{2x}-1))

calc67x

New member
Joined
Feb 6, 2018
Messages
25
I had a math problem that I need help with. I already have the answer but I can't arrive at the explanation.

It is the following:

P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/(K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1))
limx -- infinity

They want us to take the limit going to infinity. K and P[SUB]0 [/SUB]are constants.
I would think that since e[SUP]2x [/SUP]would most likely be infinity, that infinity multiplied by the constants would be infinity in the numerator.
In the denominator, there should be infinity + constant.
However, the book says the answer is K and I don't know why this would be the case.
Do any experts out there know why the answer would be K?

Thanks much for any help at understanding.
 
Last edited:

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
2,760
P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1)
limx -- infinity

They want us to take the limit going to infinity. K and P[SUB]0 [/SUB]are constants.
I would think that since e[SUP]2x [/SUP]would most likely be infinity, that infinity multiplied by the constants would be infinity in the numerator.
In the denominator, there should be infinity + constant.
However, the book says the answer is K and I don't know why this would be the case.
Do any experts out there know why the answer would be K?
I think you omitted some essential parentheses, and meant P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/(K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1)).

What you are saying is that the numerator and denominator will both approach (not be) infinity, since infinity plus a constant is still infinity. This is an indeterminate form, so you have to use some technique to rewrite it in order to find the limit.

The standard method for this kind is the multiply numerator and denominator by e[SUP]-2x[/SUP]. Try that, and see what happens. Then tell me why that worked, and how you might think of it in the future!

(Your textbook probably has at least one example somewhere of this technique, so it should look familiar to you when you see it working.)
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,703
I had a math problem that I need help with. I already have the answer but I can't arrive at the explanation.

It is the following:

P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1)
limx -- infinity

They want us to take the limit going to infinity. K and P[SUB]0 [/SUB]are constants.
I would think that since e[SUP]2x [/SUP]would most likely be infinity, that infinity multiplied by the constants would be infinity in the numerator.
In the denominator, there should be infinity + constant.
However, the book says the answer is K and I don't know why this would be the case.
Do any experts out there know why the answer would be K?

Thanks much for any help at understanding.
For starters, don't get in the habit of multiplying things by infinity. That's not a thing. It's a limit. Think of it as Increases Without Bound.

Second, be more careful with your notation. Symbols mean things. a / b + c is NOT the same as a / (b+c).

Third, why not run some experiments? I started with \(\displaystyle P_{0} = 1;and\;K =3\)

\(\displaystyle \dfrac{3*100}{3+(100-1)} = \dfrac{300}{102} = 2.9411\)

\(\displaystyle \dfrac{3*1000}{3+(1000-1)} = \dfrac{3000}{1002} = 2.9940\)

\(\displaystyle \dfrac{3*10000}{3+(10000-1)} = \dfrac{30000}{10002} = 2.9994\)

This is not a proof, by any means, but can you see what is going on? As the changing value increases, 100 ==> 1000 ==> 10000, the tiny '3' and the tiny '1' in the denominator start to mean less and less.

It often helps to rewrite a little. Example: Dividing numerator and denominator SEPARATEY by x, gives a clue: \(\displaystyle \dfrac{3\cdot x}{3 + x} = \dfrac{3}{(3/x) + 1}\) -- Is it easier to imagine what this does as x increase without bound?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,089
I had a math problem that I need help with. I already have the answer but I can't arrive at the explanation.

It is the following:

P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/(K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1))
limx -- infinity

They want us to take the limit going to infinity. K and P[SUB]0 [/SUB]are constants.
I would think that since e[SUP]2x [/SUP]would most likely be infinity, that infinity multiplied by the constants would be infinity in the numerator.
In the denominator, there should be infinity + constant.
However, the book says the answer is K and I don't know why this would be the case.
Do any experts out there know why the answer would be K?

Thanks much for any help at understanding.
What would you get if you divide - both the numerator and the denominator - by e^(2x)? and then take the limit....
 

calc67x

New member
Joined
Feb 6, 2018
Messages
25
I think you omitted some essential parentheses, and meant P(x)=(KP[SUB]0[/SUB]e[SUP]2x[/SUP])/(K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1)).

What you are saying is that the numerator and denominator will both approach (not be) infinity, since infinity plus a constant is still infinity. This is an indeterminate form, so you have to use some technique to rewrite it in order to find the limit.

The standard method for this kind is the multiply numerator and denominator by e[SUP]-2x[/SUP]. Try that, and see what happens. Then tell me why that worked, and how you might think of it in the future!

(Your textbook probably has at least one example somewhere of this technique, so it should look familiar to you when you see it working.)
Dr. Peterson:
I corrected the parentheses as you suggested
It appears that if you multiply e[SUP]2x [/SUP]with e[SUP]-2x [/SUP](e to negative exp power) you would get 1.
in that case, KP[SUB]0 [/SUB]would remain in the numerator
For the denominator, it would end up as K +P[SUB]0 [/SUB]-e[SUP]-2x[/SUP] which would be K +0.
Not sure how the P[SUB]0 [/SUB]in the numerator cancels out.
Probably I am still missing something, but I appreciate the help.
 

lev888

Junior Member
Joined
Jan 16, 2018
Messages
211
For the denominator, it would end up as K +P[SUB]0 [/SUB]-e[SUP]-2x[/SUP] which would be K +0.
Not quite correct. You need to carefully write out the result of multiplication.
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,703
Not sure how the P[SUB]0 [/SUB]in the numerator cancels out.
Probably I am still missing something, but I appreciate the help.
Look at my little example. Think REALLY hard. You should see it after a sufficient time thinking on it.

Don't use the hint just to practice arithmetic. It means something. What does it mean?
 

calc67x

New member
Joined
Feb 6, 2018
Messages
25
Dr. Peterson:
I corrected the parentheses as you suggested
It appears that if you multiply e[SUP]2x [/SUP]with e[SUP]-2x [/SUP](e to negative exp power) you would get 1.
in that case, KP[SUB]0 [/SUB]would remain in the numerator
For the denominator, it would end up as K +P[SUB]0 [/SUB]-e[SUP]-2x[/SUP] which would be K +0.
Not sure how the P[SUB]0 [/SUB]in the numerator cancels out.
Probably I am still missing something, but I appreciate the help.
Actually, I did notice in the denominator, we have K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1)
This would be K +P[SUB]0[/SUB]*e[SUP]2x[/SUP]-P[SUB]0[/SUB]
when multiplied by e[SUP]-2x[/SUP] we would get 0 +P[SUB]0[/SUB]*1-P[SUB]0[/SUB]*0 = P[SUB]0[/SUB]
Then the numerator would be KP[SUB]0 [/SUB]and the denominator would be P[SUB]0
[/SUB]So that would leave K. Does this make sense?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,176
Dr. Peterson:
I corrected the parentheses as you suggested
It appears that if you multiply e[SUP]2x [/SUP]with e[SUP]-2x [/SUP](e to negative exp power) you would get 1.
in that case, KP[SUB]0 [/SUB]would remain in the numerator
For the denominator, it would end up as K +P[SUB]0 [/SUB]-e[SUP]-2x[/SUP] which would be K +0.
Not sure how the P[SUB]0 [/SUB]in the numerator cancels out.
Probably I am still missing something, but I appreciate the help.
Here is a different way to think about this problem.

\(\displaystyle f(x) = \dfrac{KP_0e^{2x}}{K + P_0(e^{2x} - 1)} = \dfrac{K(P_0e^{2x} + K - K - P-0 + P_0)}{P_0e^{2x} + K - P_0} =\)

\(\displaystyle \dfrac{K(P_0e^{2x} + K - P_0) - K(K - P_0)}{P_0e^{2x} + K - P_0} = K - \dfrac{K(K - P_0)}{P_0e^{2x} + K - P_0}.\)

In other words, for all values of x, f(x) equals a constant minus a fraction the numerator of which is a constant and the denominator of which increases without bound as x increases without bound. What does your intuition say about what happens to that fraction as x increases without bound?
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
2,760
Dr. Peterson:
I corrected the parentheses as you suggested
It appears that if you multiply e[SUP]2x [/SUP]with e[SUP]-2x [/SUP](e to negative exp power) you would get 1.
in that case, KP[SUB]0 [/SUB]would remain in the numerator
For the denominator, it would end up as K +P[SUB]0 [/SUB]-e[SUP]-2x[/SUP] which would be K +0.
Not sure how the P[SUB]0 [/SUB]in the numerator cancels out.
Probably I am still missing something, but I appreciate the help.
You've made a mistake in distributing.

If you multiply the denominator, K+P[SUB]0[/SUB](e[SUP]2x[/SUP]-1), by e[SUP]-2x[/SUP], you have to multiply both terms, K and P[SUB]0[/SUB](e[SUP]2x[/SUP]-1); and then you have to distribute e[SUP]-2x[/SUP] P[SUB]0[/SUB] over both terms inside the parentheses.

Try again, being very careful -- take several small steps.
 

calc67x

New member
Joined
Feb 6, 2018
Messages
25
finally got it!

I tried to post earlier but apparently did not press the correct button.
I did finally notice that e-[SUP]2x[/SUP](K+P[SUB]0[/SUB](e[SUP]2x [/SUP]-1))would equal 0+P[SUB]0[/SUB]*1 -0.
This would equal KP[SUB]0 [/SUB]in the numerator and P[SUB]0 [/SUB]in denominator.
So the answer is K

Thanks for everyone's help. I will check into the concepts of the other posts as well.
What is the technique of multiplying that way called? It is almost like rationalizing the numerator?
 
Top