I don't quite understand what u mean by your statement, could you elaborate?
and for those who wanted to see how far I got
A little more detail than tkhunny initially gave: First we don't have to show for all delta, just that there exists one. So, if we can show that if x is between 2.9 and 3.1 exclusive [
EDIT: I didn't catch it at first either, just assumed an interval (2,4) was ok. So reading tkhunny's next post, I changed the interval.] If we knew the minimum, call it A, of |x-13| in that interval we could say
\(\displaystyle |\,{\frac{x-3}{x-13}}\,\,\, |\lt\, 33\, \frac{|x-3|}{A}\)
thus
\(\displaystyle {|x-3|}\, \lt\,\, \dfrac{A}{33}\,\, \epsilon\)
That is not quite sufficient. Consider what delta you would choose if epsilon were very large, i.e. if epsilon were 3*10
6, the above would give a delta of approximately 10
6. If all we are going to do is restrict |x-3| to be less than 10
6 [or even 20], our function would become unbounded on that interval.