Find the appropriate delta in terms of epsilon using the precise definition of a limit. Given the limit: lim (x→ -6) 4-3x=22. Show all steps.

Answer 1

Hi,

lim (x→ -6) 4-3x=22. Show all steps.

lim_x->-6 ( 4-3x) = 22

every | x + 6 | < δ -----> then | 4-3x - 22| < ε

| -18 -3x | < ε

factor of -3--------------------- > | -3(6 + x)| < ε

use |ab| = |a|.|b|---------------> |-3| | 6 + x | < ε

|-3| = 3 --------------------------> 3 | 6 + x | < ε

both side divide by 3---------> | 6 + x | < ε/3

NOW:

every | x + 6 | < δ then | 6 + x | < ε/3

as a result δ < ε/3

I hope it is useful,

Minoo