Epsilon delta proof of the square root function

[issacnewton]

Homework Statement

Prove that
$$ \lim_{x\to a} \sqrt{x} = \sqrt{a} $$
using $\varepsilon-\delta$ method. We are given that $a > 0$.

Relevant Equations

epsilon delta definition of a limit

Let $\varepsilon > 0$ be arbitrary. Now define $\delta = \text{min}\{\frac{a}{2}, \varepsilon \sqrt{a}\}$. Now since $a>0$, we can deduce that $\delta > 0$. Now assume the following
$$ 0< |x-a| < \delta $$
From this, it follows that $0 < |x-a| < \frac{a}{2}$ and $0 < |x-a| < \varepsilon \sqrt{a}$. We have to prove that
$$ | \sqrt{x} - \sqrt{a} | < \varepsilon $$
From $0 < |x-a| < \frac{a}{2}$, we can deduce that $0 < \frac{a}{2} < x < \frac{3a}{2}$. From here it follows that
$$ 0 < \sqrt{\frac{a}{2}} < \sqrt{x} < \sqrt{\frac{3a}{2}} $$
$$\therefore 0< \sqrt{a} + \sqrt{\frac{a}{2}} < \sqrt{x} + \sqrt{a} $$
$$\Rightarrow \frac{1}{\sqrt{x} + \sqrt{a} } < \frac{1}{\sqrt{a}(1+\frac{1}{\sqrt{2}})} $$
Now since $\sqrt{x} + \sqrt{a} > 0$, we have that $\sqrt{x} + \sqrt{a} = | \sqrt{x} + \sqrt{a} |$. So the inequality becomes as follows
$$\Rightarrow \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}(1+\frac{1}{\sqrt{2}})} < \frac{1}{\sqrt{a}}$$
$$\Rightarrow \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
Now, we can write the $| \sqrt{x} - \sqrt{a} |$ as follows
$$ | \sqrt{x} - \sqrt{a} | = \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} | } $$
We had deduced that
$$ \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
Since $0 < |x-a|$, it follows that
$$ \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} |} < \frac{|x-a|}{\sqrt{a}}$$
But with the choice of $\delta$, we have done, we have that $0 < |x-a| < \varepsilon \sqrt{a}$. Using this, we can deduce that
$$ \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} |} < \frac{|x-a|}{\sqrt{a}} < \varepsilon$$
$$\Longrightarrow | \sqrt{x} - \sqrt{a} | < \varepsilon $$

Since $\varepsilon > 0$ was arbitrary, this proves that
$$\lim_{x\to a} \sqrt{x} = \sqrt{a} \text{ if } a> 0$$

Is my proof reasonable ?
Thanks $:)$

 

[member 587159]

Problem Statement: Prove that
$$ \lim_{x\to a} \sqrt{x} = \sqrt{a} $$
using $\varepsilon-\delta$ method. We are given that $a > 0$.
Relevant Equations: epsilon delta definition of a limit

Let $\varepsilon > 0$ be arbitrary. Now define $\delta = \text{min}\{\frac{a}{2}, \varepsilon \sqrt{a}\}$. Now since $a>0$, we can deduce that $\delta > 0$. Now assume the following
$$ 0< |x-a| < \delta $$
From this, it follows that $0 < |x-a| < \frac{a}{2}$ and $0 < |x-a| < \varepsilon \sqrt{a}$. We have to prove that
$$ | \sqrt{x} - \sqrt{a} | < \varepsilon $$
From $0 < |x-a| < \frac{a}{2}$, we can deduce that $0 < \frac{a}{2} < x < \frac{3a}{2}$. From here it follows that
$$ 0 < \sqrt{\frac{a}{2}} < \sqrt{x} < \sqrt{\frac{3a}{2}} $$
$$\therefore 0< \sqrt{a} + \sqrt{\frac{a}{2}} < \sqrt{x} + \sqrt{a} $$
$$\Rightarrow \frac{1}{\sqrt{x} + \sqrt{a} } < \frac{1}{\sqrt{a}(1+\frac{1}{\sqrt{2}})} $$
Now since $\sqrt{x} + \sqrt{a} > 0$, we have that $\sqrt{x} + \sqrt{a} = | \sqrt{x} + \sqrt{a} |$. So the inequality becomes as follows
$$\Rightarrow \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}(1+\frac{1}{\sqrt{2}})} < \frac{1}{\sqrt{a}}$$
$$\Rightarrow \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
Now, we can write the $| \sqrt{x} - \sqrt{a} |$ as follows
$$ | \sqrt{x} - \sqrt{a} | = \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} | } $$
We had deduced that
$$ \frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
Since $0 < |x-a|$, it follows that
$$ \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} |} < \frac{|x-a|}{\sqrt{a}}$$
But with the choice of $\delta$, we have done, we have that $0 < |x-a| < \varepsilon \sqrt{a}$. Using this, we can deduce that
$$ \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} |} < \frac{|x-a|}{\sqrt{a}} < \varepsilon$$
$$\Longrightarrow | \sqrt{x} - \sqrt{a} | < \varepsilon $$

Since $\varepsilon > 0$ was arbitrary, this proves that
$$\lim_{x\to a} \sqrt{x} = \sqrt{a} \text{ if } a> 0$$

Is my proof reasonable ?
Thanks $:)$

On first sight, it looks perfectly fine! Note that you used that $\sqrt{\cdot}$ is a strictly increasing function. If you proved this earlier, your proof is ok.

 

[vela]

You can simplify the proof a bit by noting that because $\sqrt x > 0$, you have $\sqrt x + \sqrt a > \sqrt a$; hence, it follows that
$$\frac 1{\sqrt x + \sqrt a} < \frac 1{\sqrt a}.$$

 

[issacnewton]

yes, so if $x> 0$, then there is another theorem which guarantees that there is a unique positive square root of $x$ , such that $\sqrt{x} > 0$. So with this, we can go ahead as you guys pointed out.
Thanks