Hello,I need help in proving :
(Note a is totally different than A )
If Limit F(x) x==>a = L (equals L ) and A<L then for an 'x' which is very close to a we get that F(x)>A
Any ideads?
Thanks
Aha I see,but why its not correct to choose L - A? ;
0<|x-a|<dlta ====> |F(x)- L|< e
by |y|<m >> -m<y<m
We get ; -e<F(x)-L<e
Now e=L -A
so ==> F(x) - L > -e = -(L -A)=A - L
/ + (-L)
===> F(x)>A- L + L = A ==> F(x)> A!
Not correct ?(it is given that L -A >0,so the value of e is positive! and it fits with the requirements! )
Thank you