For f(x) to be continuous at x=2, the following 3 conditions must be met:
1- f(2) exists
2 - limx→2f(x) exists
3- limx→2f(x)=f(2)
Give a function of f (either with symbols or a graph) which satisfies the given combination of (1,2, and 3)
(a) 1 and 2 but not 3
(b) 1 but neither 2 or 3
(c) 2 but neither 1 or 3
(d) neither 1, 2, or 3
This is really just a matter of thinking about what the question is asking. The math is easy once you grasp that.
For part a), the function is defined, and the limit exists (conditions $1$ and $2$) but it has the "wrong value" at $2$ (condition $3.$) So, take any continuous function and redefine it at $2.$ For example, $$f(x)=\cases{0, &$x\neq2$\\1,&$x=2$}$$
For part b), $f(2)$ is defined, but the limit doesn't exits. Well, if the limit doesn't even exist, it clearly can't equal $f(2),$ so we get that condition $3$ fails for free. We just need a function where the limit at $2$ doesn't exist, and then we can define $f(2)$ however we like. I'll leave it to you to think of an example.
Once you've done part b), it should be easy to do part c).