suppose e>0 b>0 so if |x-a|<b >>> |x-a|<|x|<b |f(x)-f(a)|<e >>> |f(x)-f(a)|<|f(x)|<e from that step they do write same things all over again and conclude that. lim (f(x))=f(a) x->infinity how to get from the step i showed to this conclution?
Follow Math Help Forum on Facebook and Google+
Originally Posted by transgalactic suppose e>0 b>0 so if |x-a|<b >>> |x-a|<|x|<b |f(x)-f(a)|<e >>> |f(x)-f(a)|<|f(x)|<e from that step they do write same things all over again and conclude that. lim (f(x))=f(a) x->infinity how to get from the step i showed to this conclution? There must be something missing from your hypothesis. Why does Counter example and What exactly is the context of this?
i ment that if |x|<b then this is definitely true |x-a|<b
Originally Posted by transgalactic i ment that if |x|<b then this is definitely true |x-a|<b This is only true if and are of opposite signs... is one example and is another.
View Tag Cloud