Each piece of this function is continuous on its respective domain (because all polynomials are continuous functions), meaning
• 2 - x exists for all x < -1
• x exists for all -1 ≤ x < 1
• (x - 1)² exists for all x ≥ 1
So this really just leaves the points where the pieces are split up, i.e. x = -1 and x = 1. At both of these points, the two-sided limit exists as long as the one-side limits from both sides exist and are equal to one another.
At x = -1, as I said in my comment, you have
![\displaystyle \lim_{x\to-1^-}f(x) = \lim_{x\to-1}(2-x) = 2-(-1) = 3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto-1%5E-%7Df%28x%29%20%3D%20%5Clim_%7Bx%5Cto-1%7D%282-x%29%20%3D%202-%28-1%29%20%3D%203)
while
![\displaystyle \lim_{x\to-1^+}f(x) = \lim_{x\to-1}x = -1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto-1%5E%2B%7Df%28x%29%20%3D%20%5Clim_%7Bx%5Cto-1%7Dx%20%3D%20-1)
But -1 ≠ 3, so the two-sided limit
![\displaystyle\lim_{x\to-1}f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto-1%7Df%28x%29)
does not exist. So a = -1 is one of the points you would list.
At x = 1, we have
![\displaystyle \lim_{x\to1^-}f(x) = \lim_{x\to1}x = 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto1%5E-%7Df%28x%29%20%3D%20%5Clim_%7Bx%5Cto1%7Dx%20%3D%201)
while
![\displaystyle \lim_{x\to1^+}f(x) = \lim_{x\to1}(x-1)^2 = (1-1)^2 = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto1%5E%2B%7Df%28x%29%20%3D%20%5Clim_%7Bx%5Cto1%7D%28x-1%29%5E2%20%3D%20%281-1%29%5E2%20%3D%200)
and again the one-sided limits don't match, so this two-sided limit also does not exist, making a = 1 the other answer.