Given $\epsilon>0$, there exist $M$, such that $\forall x \geq M$ which leads to $$\left| \frac{x^2}{2x^2 - 1} - \frac{1}{2}\right| < \epsilon$$ Simplify. $$\left|\frac{1}{4x^2 - 2}\right|<\epsilon$$ Since x going to infinity, we may drop absolute value sign, we get $$\sqrt{\frac{1+2\epsilon}{4\epsilon}} < x$$
Now I'm stepping in something wrong. Let $$M =\sqrt{\frac{1+2\epsilon}{4\epsilon}}$$
Which should mean that any $x>M$ will suffice for definition of a limit. But if I plug in $M$ into $$\left|\frac{x^2}{2x^2 - 1} - \frac{1}{2}\right| < \epsilon$$ I'll get $\epsilon < \epsilon$. But if $x$ is bigger than $M$, result will be less than $\epsilon$, because numenator is smaller than denominator. I sort of proved it(?) , but something in my logic is wrong. I suppose my suggestion should have strict inequality? $\forall x > M$
We can also simplify noting that eventually
$$\left|\frac{1}{4x^2 - 2}\right|<\frac1{x^2}$$
and then for $\frac1{x^2}\le \varepsilon \iff x\ge M=\frac1{\sqrt \varepsilon}$
$$\left|\frac{1}{4x^2 - 2}\right|<\varepsilon$$