Limit proof by definition problem!

i need to prove this limit:

(2n^3 -3n^2 -7n +4)/( 5n^3 - 6n^2 +2n -5 ) = 2/5

i need to prove this by using the epsilon and definition. there is a specific method to deal with
these proofs?

This is actually a limit at infinity
namely

.........2n³ - 3n² - 7n + 4
lim...▬▬▬▬▬▬▬▬ = 2/5 {we wish to prove this}
n→∞..5n³ - 6n² + 2n -5


lim....f(x) = L
n→∞


If for each ε>0 {no matter how small}

there EXISTS an N ≥ 0

for which it can be shown that

|f(x) - L| < ε whenever

n > N .

Make sure you understand that small n is a variable that increments
to infinity and that large N is a constant for any FIXED VALUE OF ε

The key to this proof is to suppose that the epsilon{ε} exists then
to find an expression for N {that is capital N} that is a function of ε


For this example we SUPPOSE ε exists

then



|f(x) - L| < ε

that is


║.2n³ - 3n² - 7n + 4............2.....║
║.▬▬▬▬▬▬▬▬ ▬...▬▬ .║ < ε
║5n³ - 6n² + 2n -5...............5.....║

divide numerator and denominator of first
term in absolute value symbols by highest
power in denominator , here n cubed


║.2 - 3/n - 7/n² + 4/n³.........2.....║
║.▬▬▬▬▬▬▬▬ ▬...▬▬ .║ < ε
║5 - 6/n + 2/n² -5/n³............5.....║

next, for large positive n since n² and n³
are bigger their reciprocals are smaller
so we know the above left expression is less than ε

if


â•‘.2 - 3/n - 7/n + 4/n.........2.....â•‘
║.▬▬▬▬▬▬▬▬ ▬...▬ .║ < ε
â•‘5 - 6/n + 2/n -5/n............5.....â•‘

so we actually choose ε to satisfy above
which becomes

â•‘.2 - 6/n ...........2....â•‘
║.▬▬▬ ▬...▬▬ .║ < ε
â•‘5 - 9/n ............5.....â•‘


so
â•‘5(2 - 6/n) .......2(5 - 9/n)....â•‘
║.▬▬▬ ▬...▬▬▬▬▬ .║ < ε
â•‘5(5 - 9/n) ......5(5 - 9/n).....â•‘

thus
so
â•‘10 - 30/n) ......10 - 18/n)....â•‘
║.▬▬▬ ▬...▬▬▬▬▬ .║ < ε
â•‘5(5 - 9/n) ......5(5 - 9/n).....â•‘


so
â•‘- 12/n..........â•‘
║▬▬▬▬▬.║ < ε
â•‘25 - 45/n ..â•‘


Finish from here by solving for an n that is
a function of ε