5
Answer is 5.  Factor numerator which will allow you to cancel out an (x-2) from top and bottom leaving you with x+3. Then plug in your limit of x approaching 2 which will give you 2+3 resulting in 5.
It is indeed 5. The intuition is that -- you're not talking about the point x=2 itself. You're talking about the limit as the function approaches that point.

And since it's just a removable discontinuity, the function at x = 1.999999 or 2.0000001 is really close to five. In fact, you could get as close as you want to 5 and still not be dividibg by zero.

Hence, the *limit* as x *approaches* 2 is 5.
The expression inside the limit evaluated at 2 is undefined, however the limit itself is 5, meaning that if you are slightly away from x=2 but not actually on that point, the closer you get to x=2, the closer the expression gets to 5.

If you are at x=2.00000000000000000000001, then the expression will be very close to 5. If you add more zeroes, it will get even closer to 5. The actual point itself doesn't matter, we are only talking about points in the nearby vicinity.
It’s 5. The function *approaches* 5 as x approaches 2. It’s undefined when it *is* 2.
Also, after factoring and cancelling, the new function may be referred to as an equivalent function.