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Math Help - Continuity of f(x)

  1. #1
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    Continuity of f(x)

    Being new to continuity, how do I use the definition of continuity to show that

    f(x) = sqrt(x) is continuous on the domain of f?

    Any help would be very appreciated!
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  3. #3
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    Re: Continuity of f(x)

    Recap the definition of continuity: The function f:I\to\mathbb R is said to be continuous at the point x_0\in I iff for all \varepsilon>0, there exists \delta>0 such that for all x\in I, \left|x-x_0\right|<\delta\Rightarrow\left|f(x)-f(x_0)\right|<\varepsilon, and f is said to be continuous on I iff it is continuous at every point in I.

    For f(x)=\sqrt x, I=\left\{x\in\mathbb R:x\geqslant0\right\}.

    First show that f(x)=\sqrt x, is continuous at x_0=0. Given \varepsilon>0, take \delta=\varepsilon^2. Then for all x\geqslant0, |x-0|<\delta \implies |x|<\varepsilon^2 \implies \left|\sqrt x\right|=\left|\sqrt x-\sqrt0\right|<\varepsilon.

    Now show that f is continuous at x_0>0. Given \varepsilon>0, take \delta=\varepsilon\sqrt x_0. Then \left|x-x_0\right|<\delta \implies \left|\sqrt x-\sqrt{x_0}\right|\left|\sqrt x+\sqrt{x_0}\right|<\delta \implies \left|\sqrt x-\sqrt x_0\right|<\frac{\delta}{x+\sqrt{x_0}} \leqslant\frac{\delta}{\sqrt{x_0}}=\varepsilon.
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