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Do I have to get both $x\to2^+$ and $x\to2^-$? Then how can I get those limits? If $x\to2^+,x=2+h$ do I have to pull it in that formula? (ahh, I can't describe well ;( understand me I'm Korean) I can solve the question by drawing a graph, so please don't answer it. I really want to know how to solve it. And the answer is 6 (not 7).


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Does it help you to note that $-x^2+4x+3=-(x-2)^2+7$? – Gerry Myerson Jul 27 '12 at 5:52
For future reference: writing your entire post in capital letters with surplus exclamation points and question marks is the internet equivalent of waltzing into your professor's office and then screaming your lungs out until everyone's ears are bleeding. That isn't the best way to ask for help, is it? :) – anon Jul 27 '12 at 5:55
Note that $\lfloor 7 - (x-2)^2 \rfloor = \lfloor 7 -(2 \pm h - 2)^2 \rfloor = \lfloor 7 - h^2 \rfloor = 6.$ – user2468 Jul 27 '12 at 6:10

Hint: $-x^2+4x+3=7-(x-2)^2$; therefore, no matter how $x$ approaches $2$, you have $-x^2+4x+3<7$.

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Recalling that $$ [x+n]=[x] + n \quad (n \in \mathbb{N} ) $$ then we rearrange: $${\lim_{x\to2}[-x^2+4x+3]=\lim_{x\to2}[-(x-2)^2+7]=\lim_{x\to2}[-(x-2)^2]+7 =7+\lim_{x\to2}[-(x-2)^2]}$$ Now, we can focus exclusively on the limit: $$x\to2 ⇒0<|x-2|<\delta⇒0<(x-2)^2<\delta^2<\delta$$ (the square is a positive infinitesimal). Thus $$\lim_{x\to2}[-(x-2)^2]=-1$$ And finally $$\lim_{x\to2}[-x^2+4x+3]=7-1=6$$

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