The function Abs[x - 2]/(x^2 - 5x + 6) is not defined for x=2. Is there a value that could be given for f(2) that would make the function continuous at 2?
Give reasons for your answer.
The function Abs[x - 2]/(x^2 - 5x + 6) is not defined for x=2. Is there a value that could be given for f(2) that would make the function continuous at 2?
Give reasons for your answer.
Maddas' point when he wrote "You tell me. Are both sides of that equation actually equal for all x?" was that you cannot cancel "x- 2" because at x= 2, that would be dividing by 0.
[tex]\frac{(x- 3)(x- 2)}{x- 2}= x- 3[/math for all x except x= 2. That's why the original function, is not defined at x= 2.
Fortunately, the definition of " " says that "Given , there exist such that if 0< , then ". That "0< |x- a|" (which many people forget to write) means that the value of f(x) at x= a has no effect whatsoever on the lim as x goes to a.
Although [tex]\frac{(x- 3)(x- 2)}{x- 2}[tex] is not defined at x= 2, it has a limit there and you can make the function continuous by defining f(2) to be that limit.