Limits that exist Simple question, I just cannot find something that explains it right out and to the point without giving a huge confusing explanation. The question that I am struggling with is to determine a limit of a function if it exists.

Find: $$\lim_{x\to2}{f(x)},$$
If
$$f(x)=\begin{cases}x^{2} & \text{if } x<2 \\ 3 & \text{if } x=2 \\ 3x-2 & \text{if } x>2\end{cases}$$

Now i worked out the limits from both sides and they both equal 4. But it does say that the limit at that point equals 3. Does this mean that the limit doesn't exist? Or does this just mean that the double sided limit exists at 4, but is discontinuous and equals 3 at that point?
Thanks in advance!
 A: The answer is $4$. The limit asks for the value of a function as it approaches some $x$ value, not the exact value. There can be a hole at $x=2$ and your answer would still be valid. 
For a more technical answer, take the following definition of a limit: $$\forall \varepsilon \gt 0 \: \exists \delta\gt 0:\forall x(0\lt|x-c|\lt\delta \Rightarrow|f(x)-L|\lt\varepsilon)$$
Then notice how the difference between any $x$ value you choose and the $c$ you are approaching must always be greater than $0$.
A: The limit from either side as $x\to2$ exists and equals $4$, so the overall limit as $x\to2$ exists and equals $4$.
Limits completely ignore the value of the function at the given point, as the definition of a limit only cares about what happens in a punctured neighborhood of that point.
There is a different notion, which is of continuity, which considers both the limit of the point and the value of the point: If the limit exists and equals the value, the function is said to be continuous there. However, if the limit doesn't exist or doesn't equal the value (as we have), the function is incontinuous at the point.
