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Thread: Limit from the right

  1. #1
    Senior Member sfspitfire23's Avatar
    Oct 2009

    Limit from the right

    Hey guys,

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

    I first took $\displaystyle |\sqrt{x}-0| \rightarrow |\sqrt{x}|\times (|\sqrt{x}-0|) \rightarrow |x-0|< |\sqrt{x}| \times \epsilon$. So, $\displaystyle \delta = |\sqrt{x}| \times\epsilon$ this right? How do I get the result to follow?
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  2. #2
    Senior Member
    Feb 2010
    (Assuming the limit is to 0+). Um, let $\displaystyle \epsilon > 0$. We must find a $\displaystyle \delta > 0$, s.t. if $\displaystyle 0<x<\delta$, $\displaystyle \sqrt{x}<\epsilon$. Since $\displaystyle \sqrt{\;\cdot\;}$ is increasing, $\displaystyle \sqrt{x} < \sqrt{\delta}$. So chuse $\displaystyle \delta \le \epsilon^2$ 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.,_or,_how_I_learned_to_stop_worrying_an d_let_delta_depend_on_x.pdf
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