Limits and indeterminate forms

Discussion in 'Math' started by Mark44, Apr 28, 2008.

  1. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
    1
    There are several indeterminate forms that show up in calculus: \frac{0}{0}, \frac{\infty}{\infty}, 0 * \infty, \infty - \infty, among others.

    Here is a limit problem that is [0 * \infty] in form.

    lim n \rightarrow \infty (n * sin(\frac{\pi}{n})
     
  2. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    The infinite series of sin (x) = x - x^3/3! + x^5/5! - ...
    So If you simplify it all out. You will get
    PI - PI^3/(3!*n^2) + PI^5/(5!*n^4) -...
    So I would expect this to approach Pi since all of the other terms go to zero.
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    I don't think so. The problem is that we have sin(pi/n) alternating between -1 and 1 regardless of what n is doing. This does not sound like converging to a limitng value to me. A fundamental theorem of real analysis says you can't have a limit if the sequence of terms approacing that limit doesn not have terms that get smaller and smaller. In short, in order for the limit to exist the sequence of terms must converge. Divergent sequences don't have limits.
     
  4. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    We are not taking the limit of sin(pi/n), we are taking the limit of n*sin(pi/n).

    ------------------
    First of all the infinite series multiplication is valid.

    If n>1 which it certainly is in the limit, the terms are getting smaller.
    PI
    PI^3/3! = 5.1677127800499700292460525111836
    PI^5/5! = 2.5501640398773454438561775836953
    PI^7/7! = 0.59926452932079207688773938354604

    Since x! grows faster than PI^x, they must continue getting smaller.

    Dividing each term by n, a positive number > 1, doesn't change that. And increasing the power of the positive number, ie. 1/n 1/n^3 1/n^5, just makes it more clearly true.

    So it converges. You can rewrite the sequence as PI - 1/n(....) where the rest of the term is known to converge to a finite number using the above logic. Then the second term goes to zero as n goes to infinity. The result is Pi.

    --------------------
    As additional proof:
    http://mathforum.org/library/drmath/view/51881.html

    States that:
    "You can see that in the limit as n -> +infinity, we'll have

    sin(pi/n) / (pi/n) -> 1"

    Not exactly the same but it invalidates the argument that because sin(pi/n) is there, it can't converge.
     
  5. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
    1
    Papabravo,
    As n gets larger, sin(pi/n) approaches 0. That should be clear, if you think about it. The denominator grows large while the numerator remains the same. At the same time, the other factor, n, is growing large, which is why this limit is an indeterminate form of the type 0 * \infty.

    Mark
     
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Mark44,

    Right, and this problem seems to be suited for L'Hospital's theorem.

    lim n->infinity n*sin(pi/n) --->d(sin(pi/n)//1/n)/dn --> -cos(pi/n)*pi/n^2//-1/n^2-->pi

    So the limit converges to pi. Ratch
     
  7. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
    1
    Yes, the limit is pi. As you have shown, this expression needs a small amount of manipulation to get it into a form suitable for L'Hopital's Rule. One factor, n, gets large without bound, while the sin(pi/n) factor gets closer to zero.

    We're taught from a young age that "zero times anything is zero," which is true enough as long as the "anything" is a finite number. With this expression, though, we have an example where that's not only untrue, but it has a very nonintuitive limit of pi.
     
  8. triggernum5

    Active Member

    May 4, 2008
    216
    0
    I'm with Mark44.. The sine expansion works out to zero.. Using the same logic as Caveman cited, infinity*0=1
     
  9. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    The problem is that people try to use this property
     lim _{n \rightarrow c}(f(n) + g(n)) = lim _{n \rightarrow c}f(n) + lim _{n \rightarrow c}g(n)

    But this property is only valid if the limits on the right side exist and are finite. However, just because this property doesn't work, it doesn't mean that the limit on the left doesn't exist or is indeterminate, it simply means another way must be used.

    The method I used is valid because, after reduction using an infinite series, the limits on the right hand side fit the rule. They all go to zero except for the first term PI, which is the answer. The reduction works because due to the order of operations, it is being applied to finite terms.
     
  10. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    My maths is a bit rusty now.

    can we not simplify it to:
    sin(x)/x where x = pi/n ?

    as n tends to infinity, x tends to zero?

    Thus answer would still be pi.
     
  11. triggernum5

    Active Member

    May 4, 2008
    216
    0
    I know I was just playing devil's advocate.. Obviously the sine function isn't complimentary to multiplication..
     
  12. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
    1
    Not the same. This original limit was n*sin(pi/n), which is equal to sin(pi/n) / (1/n). You have an extra factor of pi in the numerator of the fraction in the denominator in your simplification.

    Recca, as you point out, as n grows large, pi/n (= x) approaches 0.

    As x approaches 0, sin(x)/x approaches 1, not pi as you stated.

    Mark
     
  13. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    yes but we can multiply and divide by pi
    Edit: (I knowingly left this part in simplification. it is implied in the substitution as well)
    we substituted x= pi/n (thus having an extra pi in the numerator)
    Thus we get an extra pi which gets multiplied to the 1.
    hence, the answer.
     
  14. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
    1
    OK, that makes sense. Just in case anyone else is following this, here are the details:

    n * sin(\frac{\pi}{n})
    = \pi * \frac{n}{\pi} * sin(\frac{\pi}{n})
    = \pi * sin(\frac{\pi}{n}) \div {\frac{\pi}{n}} (I can't get the LATeX tags to render one fraction over another.)

    Now as n \rightarrow \infty, \frac{\pi}{n} \rightarrow 0, and using your substitution of x = \frac{\pi}{n}, we can get the same result by taking the limit as x \rightarrow 0 of \pi * \frac{sin(x)}{x}, which is \pi.

    It can be shown in at least a couple of ways that the limit as x \rightarrow 0 of \frac{sin(x)}{x} is 1 (e.g., L'Hopital's Rule, Maclaurin series expansion of sin(x)).
    Mark
     
  15. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    There is another common way of proving limit X\rightarrow 0 sinx/x = 1. ( the one I learned before Series expansion and L'Hospital's)
    I am feeling lazy so here is the link
     
Loading...