When does a limit function not exist??

Question

$ f(x) \begin{cases} \frac{3x-6}{x^4-4}, 0 < x < 2 \\ 0, x=2 \\ \frac{x-2}{\sqrt{3-x} -1} 2 < x < 3 \end{cases}$

For this expression, can I say that these 3 cases, the limit does not exist because when $x=2$ or as $x$ approaches $2$ from left and right , all of the 3 functions have different values (when I substitute $x=2$) thus the limit does not have a finite and unique value. Is that right to say that? If not, when does a limit not exist?

1. $\lim_{x \to 2^+}$

2. $\lim_{x \to 2^-}$

3. $\lim_{x \to 2}$

Answer 1

The value of $f(x)$ when $x=2$ has nothing to do with the limit. The left hand limit ($\frac 3{32})$ and the right hand limit ($-2$) are different and that is enough to say that the limit does not exist.

Answer 2

The limits are different when taken from different directions, which is in itself enough to say that it is undefined.

1.$\lim_{x \to 2^+} f(x) = -2$ Use LHopital

2.$\lim_{x \to 2^-} f(x) = \frac{3}{32}$ Use LHopital

Since $-2 \not = \frac{3}{32}$

Therefore, 3. is undefined.