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How can we prove that $$\lim_{x\rightarrow 3} \left ( x^{3} - 3x + 2 \right ) = 20$$ using the definition with $\epsilon$ and $\delta$?

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Let $x = 3 + \delta$, $x^3-3 x+2 = 20 + 24 \delta + 9 \delta^2 + \delta^3$. Can you complete the proof now ? – Sasha Oct 14 '11 at 19:00
It is often useful to remember a fact from algebra: If you plug a number---call it "$a$"---into a polynomial and get $0$, then the polynomial is divisible by $x-a$. In this case, that means that $(x^3-3x+2)-20$ is divisible by $x-3$. I.e. it can be factored as $(x-3)(\cdots\cdots\cdots)$ (and it's not hard to figure out what goes where those dots are). You see how that's used in Chandrasekhar's answer. – Michael Hardy Oct 14 '11 at 19:06
Maybe it can be useful to prove that for continuous functions you can just plug in the value. Then you can apply this result to your polynomial. – Jonas Teuwen Oct 14 '11 at 19:42
@Jonas: I don't see that that helps unless you already know that this function is continuous, which is essentially what is to be proved. – Michael Hardy Oct 15 '11 at 0:39
Right, but maybe he already has a theorem that states that polynomials are continuous (using the $\epsilon$-$\delta$ definition for example, and not the limit definition). – Jonas Teuwen Oct 15 '11 at 10:54

You have to show $$|x^{3}-3x + 2 -20| < \epsilon \qquad \text{whenever}\ \ \ \ |x-3| < \delta$$ $$\Longrightarrow |x^{3} -3x -18| = |(x-3)| \cdot |x^{2}+3x+6| \qquad \text{whenever}\ \ \ \ |x-3| < \delta$$

Can you do it from here.

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can you explain more? i don't understand about the x^2 + 3x + 6 – Kyris Oct 14 '11 at 19:20
@Kyris: Think about what will happen to $x^{2}+3x+6$ when $x \in (3-\delta,3+\delta)$. – user9413 Oct 14 '11 at 19:26
First specify that $\delta<1$, or something else convenient. Then $|x^2+3x+6|<34$. So if $|x-3|<\epsilon/34$ we will be OK. Finally, let $\delta=\min(1,\epsilon/34$. Remember, we are given $\epsilon$ and must produce $\delta$. The "$\min$" business was just so the $\delta$ is technically right if someone gives us a ridiculous $\epsilon$, like $\epsilon=1000$. – André Nicolas Oct 14 '11 at 19:54
Okay i'll choose my delta = epsilon/(epsilon+34). OK? – Kyris Oct 14 '11 at 20:10
@Kyris: Works well. Something like $\min(1,\epsilon/34)$ is more traditional, to concentrate attention on the $\epsilon/34$ part. But anything correct, and preferably fairly simple-looking, is good. – André Nicolas Oct 14 '11 at 21:27

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