- Thread starter paofu
- Start date

- Joined
- Apr 22, 2015

- Messages
- 2,562

By the way, we cannot cancel x on top and bottom, like you did in the last line. (We cancel only factors, not numbers being added or subtracted.)

We don't need a cancellation to see that (2 + x)/(2 + x) is always 1 (because the numerator and denominator are the same). We never get 0/0 because -2 is not in the function's domain.

- Joined
- Jan 27, 2012

- Messages
- 6,704

I get it! Thank you so much for your help !!!verycloseto -2. That means x is always less than 0, whether we approach from the left or the right. Therefore, we replace |x| with -x, and the one-sided limits are both the same.

By the way, we cannot cancel x on top and bottom, like you did in the last line. (We cancel only factors, not numbers being added or subtracted.)

We don't need a cancellation to see that (2 + x)/(2 + x) is always 1 (because the numerator and denominator are the same). We never get 0/0 because -2 is not in the function's domain.

I didnâ€™t cancel the x, but (x+1), I guess itâ€™s ok then

Have a nice day!!

Got it! Thank you!

- Joined
- Apr 22, 2015

- Messages
- 2,562

Hi. I'm not sure what you did because all I see is the line drawn through x. Did you express 2+x as (x+1)+1 in your head and then cancel the (x+1) parts? That would not be a valid cancellation, either.â€¦ I didnâ€™t cancel the x, but (x+1), I guess itâ€™s ok then â€¦

As I'd mentioned, a cancellation isn't needed.