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Epsilon delta proof of the square root function
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[QUOTE="issacnewton, post: 6207709, member: 132242"] [B]Homework Statement:[/B] Prove that $$ \lim_{x\to a} \sqrt{x} = \sqrt{a} $$ using ##\varepsilon-\delta## method. We are given that ##a > 0##. [B]Relevant Equations:[/B] 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 ## :)## [/QUOTE]
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Epsilon delta proof of the square root function
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