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$\large\lim_{x\to2}[-x^2+4x+3]~?$

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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    Note that $\lfloor 7 - (x-2)^2 \rfloor = \lfloor 7 -(2 \pm h - 2)^2 \rfloor = \lfloor 7 - h^2 \rfloor = 6.$2012-07-27

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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 $\lim_{x\to2}[-x^2+4x+3]=7-1=6$