We are proving $g(x) = x^3$ is continuous at $x_0$ arbitrary.
- My attempt:
For all $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for an arbitrary point $x_0$ and $ \lvert x – x_0 \rvert \lt \delta \implies \lvert g(x) – g(x_0) \rvert \lt \epsilon$.
We plug in $g(x) = x^3$ to get:
$ \lvert x – x_0 \rvert \lt \delta \implies \lvert x^3 – x_0^3 \rvert \lt \epsilon$.
But we recognize $\lvert x^3 – x_0^3 \rvert = \lvert x – x_0 \rvert \lvert x^2 + x_0x + x_0^2 \rvert \lt \lvert x- x_0 \rvert \lvert x^2 + 2xx_0 + x_0^2 \rvert = \lvert x – x_0 \rvert \lvert (x+ x_0)^2 \rvert \lt \epsilon$.
So we have $ \lvert x^3 – x_0^3 \rvert \lt \lvert x-x_0 \rvert \lvert x+x_0 \rvert \lvert x+x_0 \rvert $
I'm not sure how to choose the proper relationship between $\delta$ and $\epsilon$ in this case.
Thank you!
Best Answer
Working from the starting point $|x^3-x_0^3|\le|x-x_0||x+x_0|^2\le\delta(2|x_0|+\delta)^2$ would suggest choosing $\delta=\min(\frac{\epsilon}{2|x_0|+1},1)$ but one of the comment points out that you made an inequality error when changing $xx_0$ to $2xx_0$. Instead, you can use $|x|\le|x_0|+|x-x_0|$ to get
$$|x^2+xx_0+x_0^2|\le(\delta+|x_0|)^2+\delta|x_0|+|x_0|^2=\delta^2+3\delta|x_0|+2|x_0|^2,$$
so $|x^3-x_0^3|\le\delta(\delta^2+3\delta|x_0|+2|x_0|^2)$ and you can pick $\delta=\min(\dfrac\epsilon{1+3|x_0|+2|x_0|^2},1)$.
Alternatively, you could just ignore the whole mess by noting that the product of continuous functions is continuous, so $x^3=x\cdot x\cdot x$ is continuous since $x$ is (choose $\delta=\epsilon$).