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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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    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).2011-10-15

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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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    @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.2011-10-14