## anonymous one year ago check my work

1. anonymous

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

2. anonymous

looks good

3. anonymous

nice explanation

4. anonymous

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$

5. anonymous

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

6. anonymous

yep

7. anonymous

how did u get that step

8. anonymous

I took a factor of 8 out

9. anonymous

yep

10. anonymous

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

11. anonymous

rooting

12. anonymous

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

13. anonymous

ah yes, your jogging my memory now!

14. anonymous

looks good tbh

15. anonymous

thanks

16. IrishBoy123

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