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Prove that the function $f(x)=x^3+3x$ is countinous at $x=1$ by working from the $\epsilon-\delta$ definition of continuity.

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  • $\begingroup$ This looks familiar. $\endgroup$ – Did Apr 19 '13 at 9:46
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Let $\epsilon>0$ we look for $\delta>0$ s.t. if $|x-1|<\delta$ then $|f(x)-f(1)|=|x^3+3x-4|<\epsilon$

we have $$|x^3+3x-4|=|(x-1)(x^2+x+1)+3(x-1)|\leq |x-1|(|x^2|+|x|+4)$$

Choose $\delta=\min(\frac{\epsilon}{10},1)$ and since $|x|<1+\delta<2$ then $$|f(x)-f(1)|\leq |x-1|(|x^2|+|x|+4)<\frac{\epsilon}{10}\times 10=\epsilon$$

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Express $\left|f(x)-f(1)\right|=\left|x^3+3x-4\right|=\left|x-1\right|\;\left|x^2+x+4\right|<\epsilon$ and find an adequate bound of $\left|x^2+x+4\right|$ on $[0,2]$.

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