Limits Question (1 Viewer)

[frenzal_dude]

Find the limit as x --> 2

(|x^2 + x -6|)/(x-2)

If you sub in 2 you get 0, but if you factorise:

(|(x-2)(x+3)|)/(x-2) are you allowed to cancel the (x-2) and just have x+3 and then sub in 2 and it gives 5? I wasn't sure if you can cancel the x-2 because the numerator is absolute value?

 

[Trebla]

Find the limit as x --> 2

(|x^2 + x -6|)/(x-2)

If you sub in 2 you get 0, but if you factorise:

(|(x-2)(x+3)|)/(x-2) are you allowed to cancel the (x-2) and just have x+3 and then sub in 2 and it gives 5? I wasn't sure if you can cancel the x-2 because the numerator is absolute value?

First of all it isn't zero when you sub in 2. It's "0/0" which is indeterminate so you need to investigate the behaviour of limit more closely.

lim_x-->2|x² + x - 6| / (x - 2)
= lim_x-->2 |(x + 3)(x - 2)| / (x - 2)
Need to consider limit from both sides:
For x-->2⁺, the numerator expression in the absolute value is positive hence
lim_x-->2⁺ |(x + 3)(x - 2)| / (x - 2)
= lim_x-->2⁺ (x + 3)(x - 2) / (x - 2)
= lim_x-->2⁺ (x + 3)
= 5
For x-->2^-, the numerator expression in the absolute value is negative hence
lim_x-->2^- |(x + 3)(x - 2)| / (x - 2)
= lim_x-->2^- [- (x + 3)(x - 2)] / (x - 2)
= lim_x-->2^- [- (x + 3)]
= - 5

Since the limit takes two different values from both sides, we conclude that the limit does not exist.