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

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

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

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