A question about $\varepsilon$-$\delta$ definition of continuity

Question

How can I prove the $f(x)=x^{\frac{2}{3}}-1$ is continuous on $[ -1, 1 ]$. First I wrote the expression for continuity on $\varepsilon$-$\delta$ language. So, $\forall \varepsilon>0$, $\exists \delta>0$ if $x_{1}\in [-1, 1]$ and $|x_{1}-x|<\delta$, then $|f(x_{1})-f(x)|<\varepsilon$. I tried so much but I ultimately get cubic deltas so I couldn't write delta as a function of epsilon. Thanks

Answer 1

First of all, continuity is about one point. What you are trying to prove seems to me like uniform continuity, which is a different thing. Luckily, continuous functions on compact sets are uniformly continuous, so you can do both. So let's prove uniform continuity (which implies continuity).

Fix $\epsilon >0$ arbitrarily small. I say that $\exists \delta$ such that if $x_1 , x_2 \in [-1,1]$ with $|x_1 - x_2 | < \delta$ then $|f(x_1)- f(x_2 )|< \epsilon$.

We have $$|f(x_1)- f(x_2)|= |(x_1)^{2/3} - (x_2)^{2/3}|=|(x_1)^{1/3} - (x_2)^{1/3}||(x_1)^{1/3} + (x_2)^{1/3}| \leq 2 |(x_1)^{1/3} - (x_2)^{1/3}|$$

Now you can use the identity $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$ and it should do the trick. I mean that you multiply and divide by $(x_1)^{2/3} + (x_2)^{2/3} + (x_1)^{1/3}(x_2)^{1/3}$ so that on the numerator you have $x_1 - x_2$ and on the denminator something which you can easily estimate.