Epsilon Delta proof of sqrt limit

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January 13th 2010, 11:28 AM
tttcomrader

Epsilon Delta proof of sqrt limit

Prove that $\lim _{x \rightarrow 4} \sqrt {x} = 2$

Proof so far.

Given $\epsilon > 0$ , let $\delta > 0$ (or some other positive values that I haven't figure out yet)

Suppose that $\mid x-4 \mid < \delta$ , I need to show that $\mid \sqrt {x} - 2 \mid < \epsilon$

Now, I know that $\mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid$ , so that $\mid \sqrt {x} -2 \mid = \frac { \mid x-4 \mid }{ \mid \sqrt {x} + 2 \mid }$

But how would I go from here? thanks.
January 13th 2010, 11:31 AM
Drexel28

Quote:

Originally Posted by tttcomrader

Prove that $\lim _{x \rightarrow 4} \sqrt {x} = 2$

Proof so far.

Given $\epsilon > 0$ , let $\delta > 0$ (or some other positive values that I haven't figure out yet)

Suppose that $\mid x-4 \mid < \delta$ , I need to show that $\mid \sqrt {x} - 2 \mid < \epsilon$

Now, I know that $\mid \sqrt {x} -2 \mid \mid \sqrt {x} + 2 \mid = \mid x -4 \mid$ , so that $\mid \sqrt {x} -2 \mid = \frac { \mid x-4 \mid }{ \mid \sqrt {x} + 2 \mid }$

But how would I go from here? thanks.

Restrict your attention to the open ball of radius one around four so that $\frac{|x-4|}{\sqrt{x}+4}\leqslant\frac{|x-4|}{\sqrt{3}+4}$
January 13th 2010, 12:05 PM
galactus

$\lim_{x\to 4}\sqrt{x}=2$

$|\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\frac{|x-4|}{\sqrt{x}+2}$

If we restrict ${\delta}$ so that ${\delta}\leq 4$ , then

$|x-4|<4$

$0<x<8$

$0<\sqrt{x}<\sqrt{8}$

$2<\sqrt{x}+2<\sqrt{8}+2$

$\frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8} +2}$

$\frac{1}{\sqrt{x}+2}<\frac{1}{2}$

$\frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|$

Therefore, $|\sqrt{x}-2|<{\epsilon}$ if $\frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ |x-4|<2{\epsilon}$

So, ${\delta}=min(2{\epsilon},4)$

See here for a very nice paper on epsilon delta limits

http://www.mathhelpforum.com/math-he...ta-proofs.html