# Epsilon Delta proof of sqrt limit

• Jan 13th 2010, 11:28 AM
Epsilon Delta proof of sqrt limit
Prove that $\displaystyle \lim _{x \rightarrow 4} \sqrt {x} = 2$

Proof so far.

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

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

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

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

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

Proof so far.

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

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

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

But how would I go from here? thanks.

Restrict your attention to the open ball of radius one around four so that $\displaystyle \frac{|x-4|}{\sqrt{x}+4}\leqslant\frac{|x-4|}{\sqrt{3}+4}$
• Jan 13th 2010, 12:05 PM
galactus
$\displaystyle \lim_{x\to 4}\sqrt{x}=2$

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

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

$\displaystyle |x-4|<4$

$\displaystyle 0<x<8$

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

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

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

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

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

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

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

See here for a very nice paper on epsilon delta limits

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