Epsilon delta proof of the square root function

  • #1
912
19
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 ## :)##
 

Answers and Replies

  • #2
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.
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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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
$$\frac 1{\sqrt x + \sqrt a} < \frac 1{\sqrt a}.$$
 
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Likes member 587159
  • #4
912
19
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
 

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