# Epsilon delta proof of the square root function

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
$$| \sqrt{x} - \sqrt{a} | = \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} | }$$
$$\frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
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 ## :)##

PeroK

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
$$| \sqrt{x} - \sqrt{a} | = \frac{|x-a| }{ |\sqrt{x} + \sqrt{a} | }$$
$$\frac{1}{|\sqrt{x} + \sqrt{a} | } < \frac{1}{\sqrt{a}}$$
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
Staff Emeritus
$$\frac 1{\sqrt x + \sqrt a} < \frac 1{\sqrt a}.$$