Out of curiosity, I typed the equation y = 2 + 0(1/(x-2)) into wolfram alpha to find that it simplified it to y = 2. Is this correct? I assumed there would be a hole or some lack of continuity at x = 2 since I didn't think 0 * undefined = 0. Is wolfram correct?
2026-03-27 07:14:08.1774595648
Does y = 2 + 0(1/(x-2)) contain a hole? Seemingly no?
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