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Math Help - Limit from the right

  1. #1
    Senior Member sfspitfire23's Avatar
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    Limit from the right

    Hey guys,

    Show that \lim_{x \to a^+} \sqrt{x} = 0 with a delta epsilon proof.


    I first took |\sqrt{x}-0| \rightarrow |\sqrt{x}|\times (|\sqrt{x}-0|) \rightarrow |x-0|< |\sqrt{x}| \times \epsilon. So, \delta = |\sqrt{x}| \times\epsilon...is this right? How do I get the result to follow?
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  2. #2
    Senior Member
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    (Assuming the limit is to 0+). Um, let \epsilon > 0. We must find a \delta > 0, s.t. if 0<x<\delta, \sqrt{x}<\epsilon. Since \sqrt{\;\cdot\;} is increasing, \sqrt{x} < \sqrt{\delta}. So chuse \delta \le \epsilon^2 and it should follow...

    Note that delta is a function of epsilon only. You can get amusing results if it is allowed to depend on x. http://math.berkeley.edu/~critch/Pi_equals_0,_or,_how_I_learned_to_stop_worrying_an d_let_delta_depend_on_x.pdf
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