Prove that $f(x) = 2x^{2}-3$ is continuous in all of $\mathbb{R}$

Question

Prove that $f(x) = 2x^{2}-3$ is continuous in all of $\mathbb{R}$

I'd like to use $\varepsilon$-$\delta$-proof for that because I still got some troubles with it.

Let $\varepsilon > 0$, let $\delta =-x_{0}+\sqrt{x_{0}^{2}+\frac{\varepsilon}{2}}$ or $\delta =-x_{0}-\sqrt{x_{0}^{2}+\frac{\varepsilon}{2}}$ . If $|x-x_{0}|< \delta$, then

$$|2x^{2}-3-(2x_{0}^{2}-3)|=|2x^{2}-3-2x_{0}^{2}+3|$$

$$=|2x^{2}-2x_{0}^{2}|=|2(x^{2} - x_{0}^{2})|$$

$$=|2\left( (x+x_{0}) (x-x_{0}) \right)| < |2((x+x_{0})\cdot \delta)|$$

$$= |2((x+x_{0}+x_{0}-x_{0}) \cdot \delta)|= |2((2x_{0}+x-x_{0}) \cdot \delta)|$$

$$= |2((2x_{0}+\delta) \cdot \delta)|= |2(2x_{0}\delta+\delta^{2})|$$

$$|4x_{0}\delta + 2\delta^{2}| = \varepsilon$$

$\Rightarrow$

$$2\delta^{2}+4x_{0}\delta - \varepsilon = 0$$

$$\delta^{2}+2x_{0}\delta - \frac{\varepsilon}{2}=0$$

$$\delta_{1,2}= -x_{0}+-\sqrt{x_{0}^{2}+\frac{\varepsilon}{2}}$$

$\delta_{1}= -x_{0}+\sqrt{x_{0}^{2}+\frac{\varepsilon}{2}}$

$\delta_{2}= -x_{0}-\sqrt{x_{0}^{2}+\frac{\varepsilon}{2}}$

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I'd like to know if I did it right?

Answer 1

It's not necessary to solve the equation for $\delta$ "exactly", you can use rougher bounds. For example, since you can choice $\delta$, you can impose that $\delta < 1$. Then by triangle inequality

$$|4(x + x_0)(x-x_0)| < 4\delta |(x-x_0) + 2x_0| \leq 4\delta(\delta + 2|x_0|) = 4\delta^2 + 8|x_0|\delta < \delta(4 + 8|x_0|)$$

Where the las step is due to the condition $\delta <1$. Finally, choosing $\delta < \min\left \{1, \dfrac{\epsilon}{(4 + 8|x_0|)}\ \right \}$ the problem is done.