Continuity- Epsilon Delta proofs

Seems fine to me, don't see any problem with it, though of course you could just use that quotients of continuous functions with non-zero denominator are continuous (presuming you're allowed to quote that)

Consdier |x+2|<= |x|+2<$\delta$+2

Set $\delta$<1

then |x+2|<3

Hence by setting $\delta$=Min(1,3$\epsilon$/2)

As written, this makes no sense. You have yet to specify what $\delta$ is.

Looks like you've got an inequality round the wrong way here. if |x+2| < 3, then $\frac{1}{|x+2|} > 1/3$, (not < 1/3).

You would do better to observe at this point that |x+2| > 1.

As written, this makes no sense. You have yet to specify what $\delta$ is.

Looks like you've got an inequality round the wrong way here. if |x+2| < 3, then $\frac{1}{|x+2|} > 1/3$, (not < 1/3).

You would do better to observe at this point that |x+2| > 1.

Admittedly better to say that but it doesn't ruin the gist of the proof.