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Homework Help: Limits, infinity, and my calculator

  1. Sep 8, 2006 #1
    I understand that the limit of sec^2(x) as x approaches pi/2 is infinity (increasing without bound), and I understand the meaning of this in terms of the epsilon-delta definition of an infinite limit.
    I also understand why the limit of sec(x) as x approaches pi/2 doesn't exist.

    What I'm a bit "sketchy" on is why my calculator (Ti-89 Titanium) displayes "infinity" as the value for sec^2(pi/2) and "undefined" for sec(pi/2) (not when evaluating the limit, but just when evaluating the function at that value). Why aren't sec^2(pi/2) and sec(pi/2) both displayed as "undefined"? Neither sec^2(x) nor sec(x) have a defined value at pi/2, do they? The fact that in one case the limit exists doesn't seem to have any effect on the value of either function at pi/2 (my book stresses the point that the limit of f(x) as x->a doesn't necessarily mean that f(a) = L, f(a) need not even exist -- which seems to be the case here)

    I'm also confused because there is a bit of ambiguity in my mind concerning "1/0" and "infinity"

    Can anyone help me understand this...?
  2. jcsd
  3. Sep 8, 2006 #2


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    When your calculator evaluates sec(pi/2)^2 it replaces the undefined value with the limit. When your calculator evalutes sec(pi/2) there is no limit since the function goes to negative infinity on one side of pi/2 and to positive infinity on the other side, so the calculator just says undefined.

    At least, that's how the calculator works. It's only a quirk of its operation. It has nothing to do with the function's actual value at pi/2, which is not defined in either case.
  4. Sep 8, 2006 #3
    Thank you, that cleared a lot up for me. :smile:
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