If we approach 1 from the left, we're using the blue function, but if we approach 1 from the right, we're using the green function. So, we have
[tex]\displaystyle \lim_{x\to 1^-}f(x) = 0,\quad\lim_{x\to 1^+}f(x) = -1[/tex]
Since the left and right limits are different, the limit
[tex]\displaystyle \lim_{x\to 1}f(x)[/tex]
does not exist.
When we approach 0, we always use the blue function. Both halves of the blue function tend to 1 as x approaches zero. So, in this case, we have
[tex]\displaystyle \lim_{x\to 0^-}f(x) = \lim_{x\to 0^+}f(x) = 1 = \lim_{x\to 0}f(x)[/tex]
Note that the fact that, by definition, we have [tex]f(0)=2[/tex] doesn't mean that the limit is wrong, or that it doesn't exist. It simply means that the function is not continuous, because we have
[tex]\displaystyle f(x_0)\neq \lim_{x\to x_0}f(x)[/tex]
As for -2, x can approach this value only from the left, because the function is not defined between -2 and -1. So, we have
[tex]\displaystyle \lim_{x\to -2}f(x) = \lim_{x\to -2^-}f(x) 0-\infty[/tex]
The limit as x approaches 3 is similar to the one where x approaches zero: the function is not defined at x=3, but the limit from both sides approaches -1:
[tex]\displaystyle \lim_{x\to 3^-}f(x) = \lim_{x\to 3^+}f(x) = -1 = \lim_{x\to 3}f(x)[/tex]
As for the limits as x approaches infinity, we have to deduce from the graph that the funtion grows indefinitely as x grows, i.e.
[tex]\displaystyle \lim_{x\to +\infty}f(x) = +\infty[/tex]
And that the function has no limit as [tex]x\to-\infty[/tex], because it has a sinusoidal behaviour, with an ever-growing amplitude.
