Show $f(x)=\sqrt{x}$ is continuous on $[0,1]$ 
Show $f(x)=\sqrt{x}$ is continuous on $[0,1]$ 


For this question, I tried two ways to do it. 
Suppose that sequence $\{x_n\}$ converges to $x_0\in[0,1]$ and $x_n\in[0,1]$ for all $n$. Then we have:
$$\lim\limits_{n\rightarrow\infty}f(x_n)=\lim\limits_{n\rightarrow\infty}\sqrt{x_n}=\sqrt{x_0}=f(x_0)$$
Therefore, $f(x)$ is continuous on $[0,1]$.
Let $\epsilon>0$ and $x_0\in[0,1]$, then there exists a $\delta=2\epsilon$ such that $|x-x_0|<\delta$ with for all $x\in[0,1]$. And for all $x,x_0\in[0,1]$, $\sup\{\sqrt{x}+\sqrt{x_0}\}=2$. Then:
$$|f(x)-f(x_0)|=|\sqrt{x}-\sqrt{x_0}|=\left|\frac{x-x_0}{\sqrt{x}+\sqrt{x_0}}\right|\leq|x-x_0|\left|\frac{1}{\sqrt{x}+\sqrt{x_0}}\right|<\delta/2=\epsilon$$
Thus, $f(x):[0,1]\rightarrow\mathbb{R}$ is continuous.

can someone check these two solution right or not? Thanks
 A: Theorem: If $f$ is continuous and one-to-one mapping of compact metric space $X$ onto metric space $Y$. Then inverse mapping $f^{-1}$ defined $f^{-1}(f(x))=x$ is also continuous.
It's easy to see that $f(x)=x^2$ implies to all condition of above theorem. $X=[0,1]$ is compact. Hence $f^{-1}=\sqrt{x}$ is continuous.
A: The first prove is incorrect  because your aim is to show that $f(x)=\sqrt(x)$ is continuous, so when taking the limit  $\lim_{n \rightarrow \infty} f(x_n)= \lim_{n \rightarrow \infty} \sqrt{x_n}$ you are not allowed to write it equals to  $ \lim_{n \rightarrow \infty} \sqrt{x_0}$. In fact what you did is   writing the limit as  $ \sqrt{\lim_{n \rightarrow \infty} x_n}= \sqrt{x_0}$, and as long as you didn't show that $f$, which is the function of square root, this function is continuous, you can't use this property of limits. 
For the second one, to show that $f$ is continuous on  $[0,1]$, you are going to show $f$ is continuous at every  $x_0 \in  [0,1]$. For that, let  $\epsilon  >0$, you have to fid   $\delta$ such that , for any $x \in [0,1]$ with  $|x- x_0|< \delta$, we have  $|f(x)-f(x_0)|< \epsilon$.
 Your mistake was that , when taking the inverse you forget to change the sign. What I mean is that, yes  $$ \sqrt{x}+\sqrt{x_0} \leq 2$$ for all $x$, $x_0  \in [0,1]$. Then  $$ \frac{1}{\sqrt{x}+\sqrt{x_0}} \geq 2$$ 
Hopefully this makes the idea clear. You can check this and this You will find a thorough proof. 
A: I would say that the first one looks not fine to me, and the second one may somewhat over-complicate the problem. You can use the theorem only if you know the continuity in advance; but this is not the case. 
The map $f: x \mapsto \sqrt{x}$ on $[0,1]$ is continuous from the right at $0$; for if $\varepsilon > 0$, then for $0 < x < \varepsilon^{2}$ we have $\sqrt{x} < \varepsilon$.
We claim that $f$ is continuous on $]0,1]$, so the proof is complete. If $c \in ]0,1]$, then for $x \in ]0,1]$ we have
$$
|\sqrt{x} - \sqrt{c}| = \bigg| \frac{x-c}{\sqrt{x}+\sqrt{c}} \bigg|;
$$
we have $|x-c| < c/2$ only if $|x| > c/2$ by triangle inequality, only if $\sqrt{x} > \sqrt{c/2}$, and only if 
$$
\bigg| \frac{x-c}{\sqrt{x}+\sqrt{c}} \bigg| < \frac{|x-c|}{\sqrt{c/2} + \sqrt{c}},
$$
which is $< \varepsilon$ if in addition we have $|x-c| < (\sqrt{c/2} + \sqrt{c})\varepsilon$; but then $|x-c| < \min \{ \sqrt{c/2} + \sqrt{c}, c/2 \}$ only if 
$|\sqrt{x} - \sqrt{c}| < \varepsilon$, and we are done.
