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lim x→13 (x^2−19x+78)/(x−13.)
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\[\lim_{x \rightarrow 13} \frac{ x²-19x+78 }{ x-13 }\]
Ok I guess I'm supposed to factorise the numerator?
yes you are on the right track ....so can you factorize it?
uh.. 78/13=6 but (x-13)(x+6)=x²-12x+78
so it could be (x-13)(x+6)+7x?
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(x-13) (x-6) is the required factorization
ok so if it would be (x-13)(x-6)+7x, then (x-13) cancels out and we're left with x-6+7x=6+6x, when we plug in x=13, we get 6+78=84 ?
what is 7x doing there?
(x-13) (x-6) /(x-13)= (x-6) 13-6 = 7 so 7 is the answer
all righty now I got it... the seven was a huge mistake that's what it was
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cool :)
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