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

Question

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

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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.

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can someone check these two solution right or not? Thanks

Answer 1

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.

Answer 2

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.

Answer 3

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.