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anonymous
 one year ago
Solve
lim X2x+12
x > 3 
X3
anonymous
 one year ago
Solve lim X2x+12 x > 3  X3

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hint: \(x^2x+12=(x3)(x\mathbin{?})\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I tried to factorize and couldnt find second term and tried with synthetic division and I couldnt

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x3)(x\mathbin{\color{red}{?}})=x^23x\mathbin{\color{red}?}x+3\mathbin{\color{red}?}\] This is supposed to be equivalent to \(x^2x+12\). This means \[\begin{cases} x^2=x^2&\text{okay}\\ x=3x\mathbin{\color{red}?}x&\text{you can solve this eq. for }\mathbin{\color{red}?}...\\ 12=3\mathbin{\color{red}?}&\text{or this one}\\ \end{cases}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0@Gloria63 how do you get 1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0if you had: \[\lim_{x \rightarrow 3}\frac{x^27x+12}{x3} \text{ I would be able \to see how you got 1 } \\ x^27x+12=(x3)(x4) \\ \lim_{x \rightarrow 3}\frac{(x3)(x4)}{x3}=\lim_{x \rightarrow 3}(x4)=34=1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0but you had \[x^2x+12\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 3}\frac{f(x)}{g(x)} \\ \\ \text{ If } f(3) \text{ and } g(3)=0 \text{ then you have more work } \\ \text{ If } f(3) \neq 0 \text{ and } g(3)=0 \text{ then the limit does \not exist }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Teacher just said this is the correct answer 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.0the limit does not exist \[\lim_{x \rightarrow 3}\frac{x^2x+12}{x3} \\ \text{ since } 3^23+12=18 \text{ and } 33=0 \\ \text{ so we have } \frac{18}{0 } \\ \text{ the limit does not exist } \\ \text{ there must be a typeo if your teacher says it is 1 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=limit%28%28x%5E2x%2B12%29%2F%28x3%29%2Cx%3D3%29 here is another source if you don't believe me
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