- #1

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- 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 ## :)##

$$ 0< |x-a| < \delta $$

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 ## :)##