Limit when denominator = 0 Can someone explain me why the following is not defined:
$$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)} = \text{not defined in real numbers}$$
But this one is $-\infty$
$$\lim_{x \to 2} \frac{-1}{(x-2)^2} = -\infty$$
Both denominator = 0 but different result.. I don't understand the difference..
 A: You may be misunderstanding limit.
The followings are correct.
$$\lim_{x\to 2-}\frac{x-3}{(x-2)(x+2)}=\infty$$
$$\lim_{x\to 2+}\frac{x-3}{(x-2)(x+2)}=-\infty$$
$$\lim_{x\to 2-} \frac{-1}{(x-2)^2}=-\infty$$
$$\lim_{x\to 2+}\frac{-1}{(x-2)^2}=-\infty$$
A: Case 1:
$$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)}$$
$$= \frac{2+h-3}{(2+h-2)(2+h+2)}$$
$$= \frac{h-1}{(h)(4+h)} $$
$$= -\infty$$
Case 2:
$$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)}$$
$$= \frac{2-h-3}{(2-h-2)(2-h+2)}$$
$$= \frac{-h-1}{(-h)(4-h)} $$
$$= +\infty$$
Hence undefined.
Where as for the other example both the limits are same and are equal to $-\infty$.Hence defined in that case.
A: Let's do what @David pointed. 
When $x\to 2^+$ then $$\frac{1}{x-2}\to +\infty$$ and while $x\to 2^-$ then $$\frac{1}{x-2}\to -\infty$$ If you are not satisfied by these, take a very small value for $x$ from both sides of $2$. For $x\in 2^+$ set $x=2+ 10^{-100}$, so $$\frac{1}{x-2}=10^{100}$$ and if we set $x=2-10^{-100}$ then we have $$\frac{1}{x-2}=-10^{100}$$
A: DavidMitra is exactly right. To see this, try graphing the two functions. 
First one: 

Second one: 

