If we have $$|x-x_0| < \delta$$ Then,
$$||x| - |x_0|| < |x-x_0| < \delta$$
$$|x| - |x_0| < \delta$$
$$|x| < \delta + |x_0|$$
We need $$|x^3-x_0^3| < \epsilon$$
Thus,
$$|x^3-x_0^3| = |x-x_0|\cdot|x^2 + x_0 x + x_0^2| = |x-x_0|\cdot|x(x+x_0)+x_0^2|$$
$$< \delta(\delta^2 + 3 \delta |x_0| + 3|x_0|^2) $$
Now if I let this equal $\epsilon$ and solve for $\delta$ I will get 3 different values. I am lost on how to proceed. I do not like the method of letting $\delta = 1$ and then at the end taking the minimum of $1$ and some other derived quantity in terms of $\epsilon$. Do I have to do it that way?
Thank you!
If $\vert x-x_0 \vert<1, $ then $\vert x \vert <\vert x_0 \vert +1$ and so $$\vert x^2+x_0x+x_0^2\vert \leq \vert x \vert^2+ \vert x_0 \vert.\vert x\vert+\vert x_0 \vert^2$$ $$\hspace{5.1cm} <\color{red}{(1+\vert x_0 \vert)^2}+\vert x_0 \vert\color{red}{(1+\vert x_0 \vert)}+\vert x_0 \vert^2 $$
Therefore we choose $$\delta=\text{min}\;\Bigg(1,\frac{\varepsilon}{(1+\vert x_0 \vert)^2+\vert x_0 \vert(1+\vert x_0 \vert)+\vert x_0 \vert^2}\Bigg)$$ so that $$\vert x^3-x_0^3 \vert < \varepsilon$$