For 1 and 2:

is it the limit as v -> -2 (negative 2)?

|v-2| => (v-2) for v>= 2 and

-(v-2) for v< 2

In both cases you described, x would be smaller than 2, Since -2< 2

Then both limits would be the same

from the left: lim v-> -2- v-2/|v-2| = (v-2)/-(v-2) = -1

from the right: lim v-> -2+ v-2/|v-2| = (v-2)/-(v-2) = -1