i'm trying to prove that
the limit of (x+3) / (x²-3) is zero as x approaching ∞
so back to definition of limit
|f(x)-L|<E
|(x+3) / (x²-3) - 0|<E
(x+3) / (x²-3) <E ,and since the left side quite complicated i've come up with simple term that always bigger than (x+3) / (x²-3)
and that is (2x) / (x²/1.5) or 3/x . this, as long i take x>3 this should maintain :
(x+3) / (x²-3) <(3/x) inequality true.
so in first equality i take M=3 and it reflect x>M
then for second inequality:
(3/x)<E so x>(3/E) in other words in this inequality M= (3/E) and i know that both M is necessary to maintain inequality between
(x+3)/(x²-3) < (3/x) < E but in reality how both M affect me to choose x ? i mean in first inequality is demand me with x>3. so should i make second M follow this?
i mean second inequality (3/x) < E
x>3/E should i restrict E with 1>E so i can choose x>3 too ?
i'm sorry if it little hard to understand, please say something if you found my words was confusing , thanks
the limit of (x+3) / (x²-3) is zero as x approaching ∞
so back to definition of limit
|f(x)-L|<E
|(x+3) / (x²-3) - 0|<E
(x+3) / (x²-3) <E ,and since the left side quite complicated i've come up with simple term that always bigger than (x+3) / (x²-3)
and that is (2x) / (x²/1.5) or 3/x . this, as long i take x>3 this should maintain :
(x+3) / (x²-3) <(3/x) inequality true.
so in first equality i take M=3 and it reflect x>M
then for second inequality:
(3/x)<E so x>(3/E) in other words in this inequality M= (3/E) and i know that both M is necessary to maintain inequality between
(x+3)/(x²-3) < (3/x) < E but in reality how both M affect me to choose x ? i mean in first inequality is demand me with x>3. so should i make second M follow this?
i mean second inequality (3/x) < E
x>3/E should i restrict E with 1>E so i can choose x>3 too ?
i'm sorry if it little hard to understand, please say something if you found my words was confusing , thanks