Finding $\delta$ for a limit statement in terms of $\epsilon$

Question

For the limit statement $\lim_{x\to c}f(x)=L$, use algebra to find $\delta > 0$ in terms of $\epsilon > 0$ so that if $0 < |x - c| < \delta$, then $|f(x) - L| < \epsilon$. $$\lim_{x\to0}\left(2x^2 − 1\right) = −1$$

This is what I have so far: $$\left|\left(2x^2 -1\right) + 1 \right| < \epsilon$$ $$\left| 2x^2 \right| < \epsilon$$ $$2 \left| x^2 \right| < \epsilon$$ I can't seem to figure out the rest of it. Any help is greatly appreciated!

Answer 1

Since $|x^2| = |x|^2$, you can further manipulate your last inequality to obtain $|x| < \sqrt{\epsilon / 2}$. So, using your work, can you now find a value of $\delta$ such that if $|x-0| < \delta$ then $|(2x^2-1)+1| < \epsilon$?

Answer 2

You can not define $\epsilon$ unless you find it.

If $|x-0|<\delta$, then $|x|^2<\delta^2$. Therefore, $$|f(x)-L|\left|=\left|(2x^2 -1\right) + 1 \right|= | 2x^2|= 2|x^2| =2|x|^2<2\delta^2$$ Now choose $\delta=\sqrt\frac{{\epsilon}}{2}$. Thus $$|f(x)-L|\left|=\left(2x^2 -1\right) + 1 \right|<\epsilon$$.