I had to read that three times before I caught it. For anyone who's still scratching their head: [0,2] (0,2] [0,2] [b]([/b]0,2] [0,2] [b][u]([/b][/u]0,2] It's a weird question, but negative infinity is definitely not the answer. In order to give you a visual sense of the reason why, look at the image I linked. It shows an example of a function where you could consider one of the minimums to be negative infinity (as you approach x=2 from the left). [url=http://www.wolframalpha.com/input/?i=%287%2F4%29x-3]Here[/url] is what your function looks like. Clearly your function doesn't approach negative infinity at 0. The error with your reasoning is that, while yes you can add an infinite number of nines after the decimal, you need to realize that each nine you add is an order of magnitude less than the one before it. You'll never add up (...down?) to negative three, much less negative infinity. As a general rule of thumb, infinity won't show up as a limit or value unless the function has the potential to divide a value by zero. With your function, since the only x is in the numerator, there is no real numerical (infinity isn't a number, it's a cardinal set) value of x for which the function can be undefined. With the function in the image I linked, when x equals 2, the denominator becomes zero, and you wind up with [b](1/0)[/b]+4, which is undefined. In general, if x isn't in the denominator, you really shouldn't have to worry about infinity. Hope that helped.