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Please check my proof

Suppose we have sequence $\sqrt{x_{n}}$

We must prove that the sequence $\sqrt{x_{n}}$ converge to $\sqrt{c}$ when c is any number in domain$[0,\infty ]$

Let $\epsilon >0$ we must find N that N

$$\sqrt{x}-\sqrt{c}<\epsilon$$

$$\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c}}{(\sqrt{x}+\sqrt{c}}>\frac{x-c}{(\sqrt{x}+\sqrt{c}}<\epsilon $$

$$\frac{x-c}{(\sqrt{x}+\sqrt{c}}<\frac{x}{\sqrt{x}}<\sqrt{x}$$

$$\sqrt{x}<\epsilon \leftrightarrow x<\epsilon ^{2}$$

$$\frac{1}{\epsilon ^{2}}<\frac{1}{x}$$

Choose $N\geq \frac{1}{\epsilon ^{2}}$

Then $\frac{1}{\sqrt{n}}<\frac{1}{\sqrt{N}}<\epsilon $

Limit of sequence $\sqrt{x}$ converge to $\sqrt{c}$ for c that is number that is in domain $[0,\infty )$

Now Plug in c that in $[0,\infty ]$ in f(x)

$$f(x)=\sqrt{x}=\sqrt{c}$$

since the sequence converge to the same value of function then it is continuous on domain

Juniven Acapulco
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Lingnoi401
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  • To show that a function $f(x)$ is continuous, it should be proved that whenever $x_n$ converges to some $c$, then $f(x_n)$ converges to $f(c)$, not the opposite. – Darío G Apr 10 '17 at 09:20
  • There are plenty more answers to this question on this thread: https://math.stackexchange.com/questions/560307/prove-that-sqrtx-is-continuous-on-its-domain-0-infty. – Joshua Siktar Aug 08 '19 at 21:11

1 Answers1

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Its hard to edit your work. There are many errors on the proof you presented. Lets take the following proof.

Let $c\in[0,\infty)$ and let $(x_n)$ be a sequence in $[0,\infty)$ such that $(x_n)$ converges to $c$. Now, we need to show that the sequence $(\sqrt{x_n})$ converges to $\sqrt{c}$. Note that $x_n\geq 0$ for all $n\in\Bbb N$. This implies that $c\geq 0$.

Case 1: $c=0$. Let $\epsilon>0$. Then $\exists N\in\Bbb N$ such that if $n\geq N$ then $|x_n-0|<\epsilon^2$. Hence, if $n\geq N$ then $|\sqrt{x_n}-0|<\epsilon$.

Case 2: $c>0$. Then $\sqrt{c}>0$ and for all $n\in\Bbb N$, we have $$|\sqrt{x_n}-\sqrt{c}|=\bigg|\frac{x_n-c}{\sqrt{x_n}+\sqrt{c}}\bigg|\leq\frac{|x_n-c|}{\sqrt{c}}$$ Since $x_n\to c$, it follows that $\frac{|x_n-c|}{\sqrt{c}}\to 0$ and hence, from the Squeeze Theorem, we get $|\sqrt{x_n}-\sqrt{c}|\to 0$ and hence, $\sqrt{x_n}\to \sqrt{c}$.

Using the Sequential Criterion for Continuity, the result follows.

Juniven Acapulco
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