Prove that $\lim _{x\to x_0} \sqrt{x}=\sqrt{x_0}\;$ for $x_0>0$

Question

Prove that $\lim _{x\to x_0} \sqrt{x}=\sqrt{x_0} \;$ for $x_0>0$

for $\epsilon >0$

consider $|\sqrt{x}-\sqrt{x_0}|=|\frac{\sqrt{x}-\sqrt{x_0}}{\sqrt{x}+\sqrt{x_0}}\sqrt{x}+\sqrt{x_0}|=|\frac{x-x_0}{\sqrt{x}+\sqrt{x_0}}|$

how to we processed from here

Answer 1

Assume that $x_{0}>0$, choose $\delta=\min\{x_{0}/2,(2^{-1/2}+1)\sqrt{x_{0}}\epsilon\}$, then for $0<|x-x_{0}|<\delta$, then $|x|\geq|x_{0}|-|x-x_{0}|>|x_{0}|-\delta\geq|x_{0}|-|x_{0}|/2=|x_{0}|/2=x_{0}/2$, then $\sqrt{x}\geq\sqrt{x_{0}/2}$, then $\dfrac{1}{\sqrt{x}+\sqrt{x_{0}}}<\dfrac{1}{(2^{-1/2}+1)\sqrt{x_{0}}}$, so $\dfrac{|x-x_{0}|}{\sqrt{x}+\sqrt{x_{0}}}<\dfrac{\delta}{\sqrt{x}+\sqrt{x_{0}}}<\epsilon$.

Answer 2

Assuming that $x_0>0$, for any $\varepsilon>0$, choose a $\delta>0$ according to $\delta\le \varepsilon\sqrt{x_0}$.

Then if $0<|x-x_0|<\delta$ we have $|\sqrt{x}-\sqrt{x_0}| = \left|\frac{x-x_0}{\sqrt{x}+\sqrt{x_0}}\right|=\frac{|x-x_0|}{|\sqrt{x}+\sqrt{x_0}|}=\frac{|x-x_0|}{\sqrt{x}+\sqrt{x_0}}<\frac{\delta}{\sqrt{x}+\sqrt{x_0}}<\frac{\delta}{\sqrt{x_0}}\le\varepsilon$.

$\blacksquare$