For the first, note that e.g.:
∣x+1∣={x+1−(x+1)x+1≥0⇒x≥−1x+1<0⇒x<−1and note also that, for this limit, you need only consider values of x close to -3. And similarly for
x2−4 of course.
The same trick may also work for the other question, I haven't tried.
[edit: you probably also want to compute the limit as
limx→3+ and
limx→3− to check that they are the same]