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anonymous
 one year ago
check my work
anonymous
 one year ago
check my work

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We have to prove \[\lim_{x \rightarrow 2}x^3=8 \space \space ; \space \space \epsilon=0.001\] \[0<x2<\delta \implies x^38<0.001\] Since \[\delta>0\]\[x2=x2\]\[0<x2<\delta \implies x^38<0.001\]\[2<x<\delta+2 \implies x^38<0.001\]\[8<x^3<(\delta+2)^3 \implies x^38<0.001\]\[0<x^38<(\delta+2)^38 \implies x^38<0.001\]\[0<x^38<(\delta+2)^38 \implies x^38<0.001\] This is only true if \[(\delta+2)^38=0.001\]\[(\delta+2)^38=\pm0.001\]\[(\delta+2)^3=8\pm0.001\] \[(\delta+2)^3=8(1 \pm \frac{0.001}{8})\]\[\delta+2=2(1\pm \frac{0.001}{8})^{\frac{1}{3}}\]\[\delta+2=2(1 \pm \frac{0.001}{24})\]\[\delta+2=2 \pm \frac{0.001}{12}\]\[\delta=\pm \frac{0.001}{12}\] Since \[\delta>0\] \[\delta=\frac{0.001}{12}=0.0833 \times 10^{3}=8.33 \times 10^{5}\] @ganeshie8 @IrishBoy123

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also we can do \[f(2+\delta)=f(2)+\epsilon\]\[(2+\delta)^3=8+0.001=8.001\] \[2+\delta=2.0000833\]\[\delta=0.0000833\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\delta+2)^3=8(1 \pm \frac{0.001}{8})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did u get that step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I took a factor of 8 out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why couldn't you just go straight to cube tubing this \[(\delta+2)^3=8\pm0.001\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(1+x)^{n} \approx 1+nx \] if x is a small quantity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah yes, your jogging my memory now!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0similar approach... \(x^38<\epsilon \implies \sqrt[3]{8\epsilon } \lt x \lt \sqrt[3]{8+\epsilon }\) Binomial expansion: \(\sqrt[3]{8\epsilon } = 2\sqrt[3]{1\frac{\epsilon}{8} } \) \( = 2 [ 1+ (\frac{1}{3})(\frac{\epsilon}{8}) + \frac{1}{2!}(\frac{1}{3})(\frac{2}{3})(\frac{\epsilon}{8})^2 + \frac{1}{3!} (\frac{1}{3})(\frac{2}{3})(\frac{5}{3})(\frac{\epsilon}{8})^3 + \cdots ] \) \(= 2[1  \frac{\epsilon}{24}  \frac{1}{9}\frac{\epsilon^2}{64}  \cdots ]\) \(\sqrt[3]{8+\epsilon } = 2[1 + \frac{\epsilon}{24}  \frac{1}{9}\frac{\epsilon^2}{64} + \cdots ]\) \(\implies 2[  \frac{\epsilon}{24}  \frac{1}{9}\frac{\epsilon^2}{64}  \cdots ] \lt x2 \lt 2[ \frac{\epsilon}{24} \color{red}  \frac{1}{9}\frac{\epsilon^2}{64} + \cdots ]\) which is *not* satisfied by \(  \frac{\epsilon}{12} \lt x2 \lt \frac{\epsilon}{12} \), notice the highlighted minus sign. ie \(\delta = { \epsilon \over 12}\) does not work for this plug the actual numbers into the equation and i think you will see. i think you might need \(\delta = \frac{\epsilon}{12}  \frac{1}{9}\cdot\frac{\epsilon^2}{32}\)
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