Hi, i have this question which i am unsure whether the answer is 0 or 2.
If x tend to 2-, I know the limit is 2.
How about if x tends to 2+? Please see image:
?? 0 certainly does exist! You mean that the value of the function, at x= 2, is not 0. And that is irrelevant to the limit!
If you look carefully at the definition of limit you will see that we require $\displaystyle |f(x)- L|< \epsilon$ only if $\displaystyle 0< |x- a|< \delta$. That "0< |x- a|" means that what happens at x= a is irrelevant. It is only what happens near the point (and, here, larger) that is important.
It is very, very important to understand that the value of a limit at a point has nothing to do with the value of the function at that point!
For example, f(x)= $\displaystyle x^2- 1$ has limit 3 as x goes to 2. Now define g(x) to be $\displaystyle x^2- 1$ as long as x is NOT equal to 2 and any value you want for x= 2. No matter what value you assign at x= 2, $\displaystyle \lim_{x\to 2} g(x)= 3$.
Sometimes students get the idea that "limit" is just another way of talking about the value of a function but that's only because "continuous" functions, which are defined as functions whose limit is the same as its value, are so easy to use. The function you have here is NOT contininuous at x= 2. In a very specific sense "almost all" functions are not continuous.