Right and left hand limits. $$ \lim_{x\to2^+} \frac{x-3}{(x-2)(x+2)}=-\infty$$
But when the same limit is evaluated with $x\to2^-$ answer comes out to be (as per book) $+\infty$.
Please list the steps and tell how.
 A: $x \rightarrow 2^+ \text{ means } x>2 \text{ or } x-2>0. \text{ so what about the other factors } \\ \text{ So we have that } x>2 \text{ subtracting 3 on both sides gives } x-3>-1 \\ \text{ this means the factor } (x-3) \text{ is near } -1 \text{ for values of } x \text{ near } 2 \text{ from the right}. \\ \text{ The factor } (x-3) \text{ is therefore negative for values of } x \text{ near } 2 \text{ from the right } . \\ \text{ Now let's look at the factor } (x+2). \\ \text{ So remember } x>2 \text{ adding 2 on both sides gives } x+2>4 \text{ which means the factor } (x+2) \\ \text{ is positive since 4 is positive for values to the near right of 2 }. \\ \text{ So we have this } \\ \frac{(\text{negative})}{(\text{positive})(\text{positive})} \infty =- \infty . \\ \text{ So doing the other way } x \rightarrow 2^- \text{ we have } x<2. \text{ you give it a try}$
A: Maybe a graph will help you visualize the limits better.

Can you see now what happens when you approach $2$ from the left and what happens when you approach $2$ from the right?
