Question

check my work

Answer 1

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
\[\delta>0\]\[|x-2|=x-2\]\[0<x-2<\delta \implies |x^3-8|<0.001\]\[2<x<\delta+2 \implies |x^3-8|<0.001\]\[8<x^3<(\delta+2)^3 \implies |x^3-8|<0.001\]\[0<x^3-8<(\delta+2)^3-8 \implies |x^3-8|<0.001\]\[0<|x^3-8|<|(\delta+2)^3-8| \implies |x^3-8|<0.001\]
This is only true if
\[|(\delta+2)^3-8|=0.001\]\[(\delta+2)^3-8=\pm0.001\]\[(\delta+2)^3=8\pm0.001\]
\[(\delta+2)^3=8(1 \pm \frac{0.001}{8})\]\[\delta+2=2(1\pm \frac{0.001}{8})^{\frac{1}{3}}\]\[\delta+2=2(1 \pm \frac{0.001}{24})\]\[\delta+2=2 \pm \frac{0.001}{12}\]\[\delta=\pm \frac{0.001}{12}\]
Since
\[\delta>0\]
\[\delta=\frac{0.001}{12}=0.0833 \times 10^{-3}=8.33 \times 10^{-5}\]

@ganeshie8 @IrishBoy123

Answer 2

looks good

Answer 3

nice explanation

Answer 4

Also we can do
\[f(2+\delta)=f(2)+\epsilon\]\[(2+\delta)^3=8+0.001=8.001\]
\[2+\delta=2.0000833\]\[\delta=0.0000833\]

Answer 5

\[(\delta+2)^3=8(1 \pm \frac{0.001}{8})\]

Answer 6

yep

Answer 7

how did u get that step

Answer 8

I took a factor of 8 out

Answer 9

yep

Answer 10

why couldn't you just go straight to cube tubing this
\[(\delta+2)^3=8\pm0.001\]

Answer 11

rooting

Answer 12

\[(1+x)^{n} \approx 1+nx \]
if x is a small quantity

Answer 13

ah yes, your jogging my memory now!

Answer 14

looks good tbh

Answer 15

thanks

Answer 16

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 \(\delta = \frac{\epsilon}{12} - \frac{1}{9}\cdot\frac{\epsilon^2}{32}\)