I've been reviewing my calculus text book and came accross this problem: Prove that the function [itex]f[/itex] defined by [itex]f(x) = \sqrt{x}[/itex] is continuous if [itex]x>0[/itex]. Would anyone mind verifying (or correcting) my proof? Suggestions are welcome. Thanks!(adsbygoogle = window.adsbygoogle || []).push({});

Proof: Let [itex]\epsilon > 0[/itex] and choose [itex]\delta[/itex] such that [itex]\delta = \mathrm{min}(x_0, \epsilon\sqrt{x_0})[/itex]. Clearly if [itex]0 < |x - x_0| < \delta[/itex] then we have that [itex]0 < x < 2x_0[/itex] which proves that [itex]x > 0[/itex]. Since [itex]x > 0[/itex] we can apply a handy factoring technique and show that [itex]|x - x_0| = |\sqrt{x} - \sqrt{x_0}|(\sqrt{x} + \sqrt{x_0}) < \delta[/itex]. Using the fact that [itex]\sqrt{x} \geq 0[/itex] we know that [itex]\sqrt{x} + \sqrt{x_0} \geq \sqrt{x_0}[/itex] and consequently [itex]|\sqrt{x} - \sqrt{x_0}|\sqrt{x_0} < \epsilon\sqrt{x_0}[/itex] completing the proof.

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# Continuity problem

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