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
[tex]\displaystyle \lim_{x\to-1^-}f(x) = \lim_{x\to-1}(2-x) = 2-(-1) = 3[/tex]
while
[tex]\displaystyle \lim_{x\to-1^+}f(x) = \lim_{x\to-1}x = -1[/tex]
But -1 ≠ 3, so the two-sided limit
[tex]\displaystyle\lim_{x\to-1}f(x)[/tex]
does not exist. So a = -1 is one of the points you would list.
At x = 1, we have
[tex]\displaystyle \lim_{x\to1^-}f(x) = \lim_{x\to1}x = 1[/tex]
while
[tex]\displaystyle \lim_{x\to1^+}f(x) = \lim_{x\to1}(x-1)^2 = (1-1)^2 = 0[/tex]
and again the one-sided limits don't match, so this two-sided limit also does not exist, making a = 1 the other answer.
