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Question Number 374 by novrya last updated on 25/Jan/15

$$S\mathrm{how}\mathrm{that}$$$$\mathrm{the}\mathrm{function}f\left(x\right)=\sqrt{x}\mathrm{continuous}\mathrm{on}[0,\mathrm{\infty })$$

Answered by prakash jain last updated on 25/Dec/14

$$\mathrm{Right}\mathrm{Hand}\mathrm{Limit}=\mathrm{Left}\mathrm{Hand}\mathrm{Limit}=\mathrm{Value}$$$$\underset{h\to 0}{\lim }\sqrt{x+h}=\underset{h\to 0}{\lim }\sqrt{x-h}=\sqrt{x},\mathrm{for}x\in (0,\mathrm{\infty })$$$$\underset{h\to 0}{\lim }\sqrt{x+h}=\sqrt{x},\mathrm{for}x=0$$$$\mathrm{Hence}f\left(x\right)=\sqrt{x}\mathrm{is}\mathrm{continous}\mathrm{on}[0,\mathrm{\infty })$$

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