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Limits and Indeterminate Forms

  1. Feb 4, 2016 #1
    Is it true that every limit that takes on an indeterminate form can be evaluated?

    Is it proper to say that a limit problem has a solution if the limit does not exist?
     
  2. jcsd
  3. Feb 4, 2016 #2

    Mark44

    Staff: Mentor

    It depends on how you define "evaluated." If, when you attempt to evaluate a limit, you get an indeterminate form, there are techniques that you can use to either a) evaluate the limit (get a number), or b) say that the limit doesn't exist (which includes ##\infty## as the "value" of the limit).
    We don't say that a limit problem "has a solution." Equations and inequalities have solutions. A limit can be a) a finite number, b) unbounded, or c) not exist at all.
    ##\lim_{x \to \infty} x^2## doesn't exist, in the sense that it is unbounded. We can also say that ##\lim_{x \to \infty} x^2 = \infty##. All this means is that ##x^2## grows large without bound as x gets large.
    ##\lim_{n \to \infty} (-1)^n## doesn't exist, period, because it oscillates forever between the two values, 1 and -1.
     
  4. Feb 4, 2016 #3
    Ok, then is it appropriate to say that every indeterminate form can be simplified?
     
  5. Feb 4, 2016 #4

    Mark44

    Staff: Mentor

    Like I said, when you get an indeterminate form, there are techniques (such as L'Hopital's Rule or algebraic techniques) that you can use to evaluate the limit or say that it doesn't exist. I wouldn't call this "simplifying" the limit expression, though. L'Hopital's Rule applies only to the ##[\frac{-\infty}{\infty}]## and ##[\frac 0 0]## indeterminate forms. Other indeterminate forms, such as ##[1^{\infty}]##, require different techniques.
     
  6. Feb 4, 2016 #5

    pwsnafu

    User Avatar
    Science Advisor

    It's worth pointing out that even L'Hopital can't be applied to every ##\frac\infty\infty## indeterminate form either. For example ##\frac{x+\sin(x)}{x}## as ##x\to\infty##.
     
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