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It says Calculate the following limits:

lim x-> -3 ((1/x+3)+6/(x^2-9)) So what i did was changed x^2-9 into (x-3)(x+3) and then made that the common denominator so I got

1(x-3) + 6/(x-3)(x+3) then the (x-3)'s cancel and your left with 6/(x+3) but then I plug int he -3 and I get 6/0 which is undefined, is that the answer that it doesnt exist or am I doing something wrong? THANKS

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# 1st year calc, limits

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