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**oblixps** so i combined the fractions and got (3x^2 - |x+y|)/(3|x+y|). i found that 3 </= |x+y| </= 2δ + 3 since 0<|x-1|<δ and similarly with |y-2|. so (3x^2 - |x+y|)/(3|x+y|) </= (3x^2 + 2δ + 3)/(9). then i found that |x|-1 </= |x-1| < δ so |x|< δ + 1 and x^2 < (δ+1)^2 so (3x^2 + 2δ + 3)/(9) </= (3(δ+1)^2 + 2δ + 3)/(9). so finish the proof i just have to say that given any epsilon i can pick a delta since ε = (3(δ+1)^2 + 2δ + 3)/(9).

does that look correct? i hope i didn't make any big mistakes.