check my work

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We have to prove \[\lim_{x \rightarrow 2}x^3=8 \space \space ; \space \space \epsilon=0.001\] \[0<|x-2|<\delta \implies |x^3-8|<0.001\] Since \[\delta>0\]\[|x-2|=x-2\]\[00\] \[\delta=\frac{0.001}{12}=0.0833 \times 10^{-3}=8.33 \times 10^{-5}\] @ganeshie8 @IrishBoy123
looks good
nice explanation

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Also we can do \[f(2+\delta)=f(2)+\epsilon\]\[(2+\delta)^3=8+0.001=8.001\] \[2+\delta=2.0000833\]\[\delta=0.0000833\]
\[(\delta+2)^3=8(1 \pm \frac{0.001}{8})\]
yep
how did u get that step
I took a factor of 8 out
yep
why couldn't you just go straight to cube tubing this \[(\delta+2)^3=8\pm0.001\]
rooting
\[(1+x)^{n} \approx 1+nx \] if x is a small quantity
ah yes, your jogging my memory now!
looks good tbh
thanks
similar approach... \(|x^3-8|<\epsilon \implies \sqrt[3]{8-\epsilon } \lt x \lt \sqrt[3]{8+\epsilon }\) Binomial expansion: \(\sqrt[3]{8-\epsilon } = 2\sqrt[3]{1-\frac{\epsilon}{8} } \) \( = 2 [ 1+ (\frac{1}{3})(-\frac{\epsilon}{8}) + \frac{1}{2!}(\frac{1}{3})(-\frac{2}{3})(-\frac{\epsilon}{8})^2 + \frac{1}{3!} (\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})(-\frac{\epsilon}{8})^3 + \cdots ] \) \(= 2[1 - \frac{\epsilon}{24} - \frac{1}{9}\frac{\epsilon^2}{64} - \cdots ]\) \(\sqrt[3]{8+\epsilon } = 2[1 + \frac{\epsilon}{24} - \frac{1}{9}\frac{\epsilon^2}{64} + \cdots ]\) \(\implies 2[ - \frac{\epsilon}{24} - \frac{1}{9}\frac{\epsilon^2}{64} - \cdots ] \lt x-2 \lt 2[ \frac{\epsilon}{24} \color{red} - \frac{1}{9}\frac{\epsilon^2}{64} + \cdots ]\) which is *not* satisfied by \( - \frac{\epsilon}{12} \lt x-2 \lt \frac{\epsilon}{12} \), 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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