Question Number 374 by novrya last updated on 25/Jan/15
$${S}\mathrm{how}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)=\sqrt{{x}}\:\mathrm{continuous}\:\mathrm{on}\:\left[\mathrm{0},\infty\right) \\ $$
Answered by prakash jain last updated on 25/Dec/14
$$\mathrm{Right}\:\mathrm{Hand}\:\mathrm{Limit}=\mathrm{Left}\:\mathrm{Hand}\:\mathrm{Limit}=\mathrm{Value} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{{x}+{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{{x}−{h}}=\sqrt{{x}},\:\:\mathrm{for}\:{x}\in\left(\mathrm{0},\infty\right) \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{{x}+{h}}=\sqrt{{x}},\:\:\mathrm{for}\:{x}=\mathrm{0} \\ $$$$\mathrm{Hence}\:{f}\left({x}\right)=\sqrt{{x}}\:\mathrm{is}\:\mathrm{continous}\:\mathrm{on}\:\left[\mathrm{0},\infty\right) \\ $$