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Difference Between lim(n->inf.) n and Infinity

  1. Apr 22, 2012 #1
    Let's consider this problem: 0 × lim(n→∞) n

    If the limit were evaluated first, it would be 0 × ∞ = undefined.
    If the limit were rewritten as lim(n→∞) 0n, it would be lim(n→∞) 0 = 0.

    I cannot tell which is the correct interpretation. It seems that if we consider n to be a finite and growing value, approaching a limit, it will always equal zero. But if we consider it to be the limit itself, it seems to be undefined.

    Which is the correct interpretation? I feel as though this is a fundamental characteristic of limits that I am not understanding.
     
  2. jcsd
  3. Apr 22, 2012 #2
    The limit as n increases without bound of n diverges. That is, it does not converge to a real number, so we cannot multiply it by 0 the way we multiply real numbers by 0. When we write that the limit is ∞, we do not mean that the limit converges to a real number that we can use in operations. ∞ is a way of talking about a specific kind of divergent limiting behavior, since there are several important types.
    Thus, the first expression is an indeterminate form. We can determine it by putting the constant back into the limit: the limit as n approaches infinity of the term 0*n is 0, since 0*n = 0 for all values of n. This expression is only related to the previous expression by the fact that we can factor out 0 to get an indeterminate form. Forms are indeterminate because they may converge to more than one limit, depending on the precise expressions involved.
    Ie., the form 0x∞ cannot simply be said to be 0, because the original form might be the limit as n approaches 0 of n*1/n. This gives the indeterminate form 0x∞, but it is not 0, the limit is 1. Each case of an indeterminate form must be dealt with separately.
     
    Last edited: Apr 22, 2012
  4. Apr 22, 2012 #3
    Correct. But the way you've written it, there is only ONE interpretation, namely to take the limit first.

    If you wrote

    $$\lim_{n\to\infty}(0 * n)$$

    then you would evaluate the expression in parens first, and you'd note that for all n, 0 * n = 0, so the limit is 0.

    So yes, the grouping is significant. But what you originally wrote was NOT ambiguous ... the limit is applied only to n, so you end up with 0 * ∞, an indeterminate form.

    Another way to say the same thing is that in your example, the limit is applied to n; and in the example I wrote, the limit applies to 0 * n. So it's like anything else in symbolic math ... the meaning of an expression is a function of the specific string of symbols. Write it a different way, and you change the meaning.
     
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4
    I think I see what you mean, slider. Thanks.

    I was under the impression that lim (kx) = k lim (x), which would make my two equations equivalent. Is this not the case, and I am just thinking of most other operations in calculus that I've learned?
     
  6. Apr 22, 2012 #5
    The equality lim(kx)=k*lim(x) if and only holds if lim(x) exists (that is, it is some finite real number). That is not the case in your example.
     
  7. Apr 22, 2012 #6
    Ah, I see! Thank you for this!
     
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