- #1
issacnewton
- 1,003
- 31
- 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 ## :)##