How can we prove that $\lim_{x\rightarrow 3} \left ( x^{3} - 3x + 2 \right ) = 20$ using the definition with $\epsilon$ and $\delta$?
How to prove this limit using definition?
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calculus
limits
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0Right, 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
1 Answers
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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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0@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