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Limit of this

  1. Feb 22, 2007 #1

    ranger

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    So I've been given a differential equation that models bacterial growth, p(t), and the concentration of critical substance, q(t), whatever thats suppose to mean. I've solved both of these and found that:
    [tex]q(t) = q_0 e^{vt}[/tex]
    Where q0 is the amount critical substance at t=0. v is a constant.

    I am then asked to take the limit of q(t) as follows:
    [tex]\lim_{t \to \infty} \frac{q(t)} {1 + q(t)}[/tex]
    which comes out to be:
    [tex]\lim_{t \to \infty} \frac{ q_0 e^{vt}} {1 + (q_0 e^{vt})}[/tex]

    Its been a while since I've done limits of this sort. So how would I approach this?

    I'm just doing some exercises. Mentors can move it where they see fit
     
  2. jcsd
  3. Feb 22, 2007 #2

    StatusX

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    Roughly speaking, the exponential term gets huge while 1 stays about the same size, so the bottom begins to look more and more like the top, and so the limit is 1. More formally, multiply the top and bottome by 1/q(t), and you'll get something that should be slightly easier to deal with. Unless v<0, but since you're talking about "growth", I'm guessing it isn't.
     
  4. Feb 22, 2007 #3
    Is this limit in an indeterminate form (for certain values of v)? If so can we use l'hopital's rule on it dirrectly?
     
  5. Feb 22, 2007 #4

    ranger

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    Thanks for the quick response.

    But multiplying the top and bottom by 1/q(t) seems to put me back to where I started.
     
  6. Feb 22, 2007 #5
    If V>0 what does the numerator's limit tend to? If V>0 what does the denominator's limit tend to? Ask your self the same questions about V<0.
     
  7. Feb 22, 2007 #6

    StatusX

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    Doing that leaves 1/(1+1/q(t)). What happens to 1/q(t) when t gets very big?
     
  8. Feb 22, 2007 #7

    ranger

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    Well this is my entire problem. If I choose a value greater than zero for v and I let t get bigger and bigger, the overall value increases (inf?). The same happens for the denominator. Conceptually I guess I can see whats going to happen, but I simply forgot how to do these kinds of this with inf on paper. I think the q0 in there is also confusing me. I feel so embarrassed :shy: btw, V would always be greater than zer0 as per the problem parameters.
    The quotient gets smaller and smaller.
     
  9. Feb 22, 2007 #8
    can you see that if we look at the numerator and denominator as individual functions they each tend towards infinity for v>0? If so do you remember what l'hopital's rule says?
     
  10. Feb 22, 2007 #9
    let f(t) = q_0e^(vt)
    let g(t) = 1 + q_0e^(vt)

    l'hopital's rule says that if as t tends towards infinity f(t) tends towards infinity and g(t) tends towards infinity, then the limit as t tends towards infinity of f(t)/g(t) is the same as f'(t)/g'(t).
     
  11. Feb 22, 2007 #10

    ranger

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    OK, I think i understand what you are saying. So becuase my t is approaching inf, both the numerator and denominator will become inf, giving inf/inf which is indeterminate, so if I keep on taking the derivative, the limit would always be inf/inf, correct? I can make life easier by doing the following:
    [tex]\lim_{t \to \infty} \frac{e^{t}} {e^{t}}[/tex]
    But how can I have an indeterminate answer for a problem thats related to population growth?
     
    Last edited: Feb 22, 2007
  12. Feb 23, 2007 #11
    Hello ranger,

    multiply your fraction by [tex]\frac{e^{-vt}} {e^{-vt}}[/tex] :

    [tex]\lim_{t \to \infty} \frac{e^{-vt}} {e^{-vt}} \frac{ q_0 e^{vt}} {1 + (q_0 e^{vt})} [/tex]
     
  13. Feb 23, 2007 #12

    HallsofIvy

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    Ranger, StatusX's initial suggestion is the simplest. The crucial point is not just that t is going to infinity, but that, according to you q(t) is going to infinity.
    Dividing both numerator and denominator of
    [tex]\frac{q(t)}{1+ q(t)}[/tex]
    by q(t) gives, just as you said before
    [tex]\frac{1}{\frac{1}{q(t)}+ 1}[/tex]
    Now, as q(t) goes to infinity, again, as you said, 1/q(t) goes to 0 so that whole fraction goes to what?
     
  14. Feb 23, 2007 #13

    ranger

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    So since 1/q(t) in the denominator goes to zero, we would simply have 1/1. Which of course comes out to be 1, right?

    One more question. Taking the limit in the original form (without multiplying by 1/q(t)), we would get inf/inf. But we are sort of manipulating the problem so we dont get a indeterminate answer, which is understandable for a problem like this. But when faced with limits other than this, we would still have to do this sort of thing?
     
    Last edited: Feb 23, 2007
  15. Feb 23, 2007 #14

    HallsofIvy

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    If two functions are the same same for all x except the point at which the limit is being taken, then their limits are the same so "manipulating the problem" works. I'm not sure what you mean by "limits other than this". If there is a simpler way to find a limit (for example [itex]\lim_{x\rightarrow a} f(x)/g(x)[/itex] where f and g are continuous and g(a) is NOT 0) then you certainly don't HAVE to do it!
     
  16. Feb 23, 2007 #15
    The others' answers seem more intuitive. But in a pinch, it never hurts to L'Hopitalize a quotient (assuming, of course, that it's an indeterminate form). Do you remember the conditions of L'Hopital's Rule?
     
  17. Feb 23, 2007 #16

    ranger

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    Since this is in indeterminate form, lets see what happens when L'Hopital's rule is applied. But it seems that if I take the derivative of the numerator and denominator, I'd still get inf/inf. I am applying L'Hopital's correctly?
     
  18. Feb 23, 2007 #17

    rbj

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    it depends on whether [itex]v>0[/itex], [itex]v=0[/itex], or [itex]v<0[/itex] . is that known?
     
  19. Feb 23, 2007 #18

    ranger

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    Yup. both v and q0 are greater than zero.

    I've already found the limit to be 1. But I want to see how L'Hopital would work. Believe it or not, I dont ever recall using L'Hopital rule to evaluate a limit. But then again its be 1.5 years since calc I.
     
  20. Feb 24, 2007 #19

    VietDao29

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    L'Hopital's rule stated that:
    If [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}[/tex] is of the Indeterminate Form 0/0 or [tex]\frac{\infty}{\infty}[/tex], g′(x) is nonzero throughout some interval containing [tex]\alpha[/tex], and that the limit: [tex]\lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}[/tex] exists, then:

    [tex]\fbox{\mathop{\lim} \limits_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \mathop{\lim} \limits_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}}[/tex]​

    In the problem, since v > 0 (as you stated), we have:
    [tex]t \rightarrow +\infty \Rightarrow q(t) = q_0 e ^ {vt} \rightarrow \infty[/tex], the limit has the form [tex]\frac{\infty}{\infty}[/tex].
    By using L'Hopital's rule, we obtain:
    [tex]\lim_{t \rightarrow + \infty} \frac{q_0 e ^ {vt}}{1 + q_0 e ^ {vt}} = \lim_{t \rightarrow + \infty} \frac{(q_0 e ^ {vt})'}{(1 + q_0 e ^ {vt})'} = \lim_{t \rightarrow + \infty} \frac{vq_0 e ^ {vt}}{vq_0 e ^ {vt}} = 1[/tex]
    :)
     
    Last edited: Feb 24, 2007
  21. Feb 24, 2007 #20

    ranger

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    hehe, now I get it. Thank you VietDao29.
     
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