Proving that $f(x)=2x^2-1$ is continuous using the epsilon delta definition

Question

$f:\mathbb{R}\rightarrow \mathbb{R}$

$f(x)=2x^2-1$

Let ε>0 we have to find δ>0 such that if $x\in \mathbb{R}$ and $ \left | x-x_{0} \right |<δ$ then $\left | f(x)-f(x_{0}) \right |<ε$,

$\forall x \in \mathbb{R}$:

$\left |f(x)-f(x_{0}) \right|=$ $\left |2x^2-1-2x_{0}+1 \right|$=$\left |x^2-x_{0}^2 \right|$=$ 2 \left | x+x_{0} \right| \cdot \left| x-x_{0} \right|$<$2\left| x-x_{0}+2x_{0} \right|\cdotδ$

$\leq2(\left| x-x_0 \right|+2\left|x_0 \right|)\cdotδ$ $<2(δ+2\left|x_0 \right|)\cdot δ<ε$

Let δ=1

$2(1+2\left|x_0 \right|)<ε$

So we choose $δ=min\left\{1,2(1+2\left|x_0 \right|)\right\}$

Is this correct? Because my book suggests another solution

Answer 1

It's correct. Another suggesstion is $$\delta^2+2|x_0|\delta+|x_0|^2<\dfrac12\varepsilon+|x_0|^2$$ then $$\delta<\sqrt{\dfrac12\varepsilon+|x_0|^2}-|x_0|$$