Solve lim X2-x+12 x -> 3 --------- X-3

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Solve lim X2-x+12 x -> 3 --------- X-3

Calculus1
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Hint: \(x^2-x+12=(x-3)(x-\mathbin{?})\)
I tried to factorize and couldnt find second term and tried with synthetic division and I couldnt
what happened when you plugged in 3? if you got something/0 where the something was zero then you have more work if you have something that was not 0 then your work is done and you can say the limit does not exist

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so what is 3^2-3+12?
Answer must be -1
\[(x-3)(x-\mathbin{\color{red}{?}})=x^2-3x-\mathbin{\color{red}?}x+3\mathbin{\color{red}?}\] This is supposed to be equivalent to \(x^2-x+12\). This means \[\begin{cases} x^2=x^2&\text{okay}\\ -x=-3x-\mathbin{\color{red}?}x&\text{you can solve this eq. for }\mathbin{\color{red}?}...\\ 12=3\mathbin{\color{red}?}&\text{or this one}\\ \end{cases}\]
@Gloria63 how do you get -1?
if you had: \[\lim_{x \rightarrow 3}\frac{x^2-7x+12}{x-3} \text{ I would be able \to see how you got -1 } \\ x^2-7x+12=(x-3)(x-4) \\ \lim_{x \rightarrow 3}\frac{(x-3)(x-4)}{x-3}=\lim_{x \rightarrow 3}(x-4)=3-4=-1\]
but you had \[x^2-x+12\]
\[\lim_{x \rightarrow 3}\frac{f(x)}{g(x)} \\ \\ \text{ If } f(3) \text{ and } g(3)=0 \text{ then you have more work } \\ \text{ If } f(3) \neq 0 \text{ and } g(3)=0 \text{ then the limit does \not exist }\]
Teacher just said this is the correct answer -1
the limit does not exist \[\lim_{x \rightarrow 3}\frac{x^2-x+12}{x-3} \\ \text{ since } 3^2-3+12=18 \text{ and } 3-3=0 \\ \text{ so we have } \frac{18}{0 } \\ \text{ the limit does not exist } \\ \text{ there must be a type-o if your teacher says it is -1 }\]
http://www.wolframalpha.com/input/?i=limit%28%28x%5E2-x%2B12%29%2F%28x-3%29%2Cx%3D3%29 here is another source if you don't believe me

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