Q:

Calculating limits. Am I correct? (with working)

This is my working for the first two questions (i and ii). Does anyone know how to work out iii? Also, is my working correct? Any feedback is appreciated. I also did a rough calculation of iii and I think the answer is supposed to be 5 but I'm not too sure of the working. 
(i)
We are finding the limit as x approaches -2. According to the system of equations, we should look at where it says -2 ≤ x ≤ 3 because x is approaching -2 and -2 fits into this system.
So Ix^2 - 4I if -2 ≤ x ≤ 3 is what we're looking at. What we're finding is the limit if x approaches -2. So let's replace x with -2.
Ix^2 - 4I
I(-2)^2 - 4I
I(4) - 4I
I0I
0
lim f(x) as x app. -2 = 0.
(ii)
lim f(x) as x app. 2. This fits into Ix^2 - 4I if -2 ≤ x ≤ 3. Same reasoning as above. After it says if, it says if -2 ≤ x ≤ 3 and if we replace 2 for x here it makes sense. So now we replace 2 for x where it says Ix^2 - 4I.
Ix^2 - 4I
I(2)^2 - 4I
I(4) - 4I
I0I
0
lim f(x) as x app. 2 = 0.

A:

i) Notice that the function has two different expressions at $2^-$ and $2^+$ and we have
$$\lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}\frac{1}{x+2}=-\infty$$
and
$$\lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}|x^2-4|=0$$
ii) is correct
iii)the same method as i)  and we find
$$\lim_{x\to 3^-}f(x)=5= \lim_{x\to 3^+}f(x)\neq f(0)$$
hence we conclude that $f$ isn't continuous at $3$.

