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How to express the following limit in epsilon-delta language

  1. Nov 23, 2006 #1
    Express the following as an epsilon-delta proof (to show that it is continuous):

    [tex]\lim_{x\rightarrow - \infty}f(x) = L[/tex]

    Can I get some ideas on this one?
  2. jcsd
  3. Nov 24, 2006 #2

    matt grime

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    No one can do that question: you have failed to say what f is.
  4. Nov 24, 2006 #3


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    Just express that in terms of [itex]\epsilon[/itex], [itex]\delta[/itex] for general f and L? Then just use the definition of limit at infinity: the "standard" definition of limit, at some number a, is "Given any [itex]\epsilon> 0[/itex], there exist [itex]\delta[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex]." If you really mean "at infinity" then "[itex]|x- a|< \delta[/itex]" has to become "x sufficiently large" or "x> X" for some number X.

    The definition, then, is
    Given any [itex]\epsilon> 0[/itex], there exist a real number X such if x> X then [itex]|f(x)- L|< \epsilon[/itex].

    The reason I said 'If you really mean "at infinity"' is that you also say "to show that it is continuous". You are taking the limit as x goes to infinity and a function is never continuous "at infinity". Taking the limit as x GOES to infinity simply means "as x gets larger without bound". "Infinity" is not a specific number and f is not defined "at infinity".
    I wonder if you don't actually mean "limit as x goes to a" for some number a.
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