Please post the entire question.
i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}
ok i tried to prove the same thing with
my book says that this limit would work only if
for every N>0 we could find M>0 so for every x>M f(x)>N would work
so i need to find M for which >N would work
x>M
this is as far as i could go
there is no absolute value innequality for me to play with like you did previosly
ok but like you did previosly this is only a sketch i do for my self to find the M.
the actual proof starts with : (i tried to do it like you did)
there is N for which we pick M to be
so thus saying that x>0
what now , here i proved the limit for not for the original limit.
how i connect it to the original limit
?