the question and where i got stuck in this link:
http://img187.imageshack.us/img187/8868/44429492ix7.gif
the question and where i got stuck in this link:
http://img187.imageshack.us/img187/8868/44429492ix7.gif
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 $\displaystyle \delta =\min \left\{ 1,\frac{\epsilon }{11} \right\}.\quad\blacksquare$
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
$
did i prove it correctly
i solved it differently from the article
is that ok?
http://img224.imageshack.us/img224/9929/93418507so5.gif