# limit prooving question..

• Jan 6th 2009, 09:31 AM
transgalactic
limit prooving question..
the question and where i got stuck in this link:

http://img187.imageshack.us/img187/8868/44429492ix7.gif
• Jan 6th 2009, 10:45 AM
Krizalid
We have to prove the following statement:

$\displaystyle \left| x-5 \right|<\delta \implies \left| x^{2}-25 \right|<\epsilon .$

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Observe that $\displaystyle \left| x^{2}-25 \right|=\left| x+5 \right|\left| x-5 \right|.$ Now pick $\displaystyle \delta=1,$ and $\displaystyle \left| x+5 \right|\left| x-5 \right|\le 11\left| x-5 \right|<\epsilon .$

Hence, conditions are fulfilled if you take $\displaystyle \delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare$
• Jan 6th 2009, 11:43 AM
transgalactic
my question i about how to choose the value for M

??
• Jan 6th 2009, 11:59 AM
transgalactic
i understand that our upper bound is 11

so inorder for this to work (like your previous examples in the article you published
we need only

$\displaystyle $$\displaystyle \delta = \frac{\epsilon }{11} why to use minimum?? the previoss werenot solved like that \displaystyle$$\displaystyle \delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare$
• Jan 6th 2009, 12:18 PM
transgalactic
we get x as close as possible to 5

|x-5|<1 ==>> -1<x-5<1 so x+5<11
so M>=11
lambda=e/11

why minimum with 1 ??
• Jan 6th 2009, 02:17 PM
Krizalid
It's just any number you want.

I took $\displaystyle x\in(4,6)$ because $\displaystyle \delta=1$ is a nice guy. You could've taken other delta, a ratio, 2, whatever.
• Jan 6th 2009, 08:20 PM
transgalactic
why you didnt do this min function in the previous examples
in the article?
• Jan 6th 2009, 09:11 PM
transgalactic
usually ou ended this question by saying lambda=e/11

why lambda=min{1,e/11}

why 1? why any number i want?
why minimum?
• Jan 7th 2009, 12:19 AM
transgalactic
did i prove it correctly
i solved it differently from the article
is that ok?

http://img224.imageshack.us/img224/9929/93418507so5.gif