# Epsilon delta proof of the square root function

• issacnewton
In summary, the proof shows that for any arbitrary positive value of epsilon, we can choose a corresponding value of delta such that for any x that satisfies the given conditions, the difference between the square root of x and the square root of a is less than epsilon. This proves that the limit as x approaches a of the square root of x is equal to the square root of a.

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

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
$$\frac 1{\sqrt x + \sqrt a} < \frac 1{\sqrt a}.$$

member 587159
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

## 1. What is an epsilon delta proof?

An epsilon delta proof is a method used in mathematical analysis to rigorously prove the limit of a function. It involves choosing a small distance (epsilon) and showing that for every point within a certain distance (delta) of the limit, the function values are also within that distance from the limit.

## 2. How is an epsilon delta proof applied to the square root function?

The epsilon delta proof for the square root function involves showing that for any given epsilon (error) value, there exists a delta (distance from the limit) such that the difference between the square root function value and the limit is less than epsilon for all x values within delta of the limit.

## 3. Why is an epsilon delta proof important?

Epsilon delta proofs provide a rigorous and precise way to prove the limit of a function. This is important because it ensures that the limit is truly the value that the function approaches, and not just an estimate or approximation.

## 4. What are the key steps in an epsilon delta proof?

The key steps in an epsilon delta proof include choosing an arbitrary epsilon value, finding a corresponding delta value, showing that the function values within delta of the limit are within epsilon of the limit, and concluding that the limit is indeed the value that the function approaches.

## 5. Can an epsilon delta proof be used for any function?

Yes, an epsilon delta proof can be used for any function as long as the function is continuous at the limit point. This means that the function values approach the limit value as the input values approach the limit point.

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