<s>[INST] Proving $\lim _{x \to \infty }\frac{1 - 2x^2}{x^2 + 3}\ = -2$ I solved it using the definition of limit to the $\left|\frac{-7}{x^2 +3}\right| < \varepsilon$
Then $\varepsilon > \frac{7}{x + 3} > \frac{7}{x^2 +3}$
Then $x > \frac{7}{\varepsilon} - 3$
But my teacher said this was wrong and I have to fix it to
$x >  \sqrt{\left|\frac{7}{\varepsilon} - 3\right|}$
Can anyone please explain why am I wrong here? Thank you very much
 [/INST] We have that for $0<\varepsilon <\frac 7 3$
$$\left|\frac{1 - 2x^2}{x^2 + 3}\ -(-2)\right|<\varepsilon \iff \left|\frac{7}{x^2 + 3}\ \right|<\varepsilon \iff x^2>\frac 7 \varepsilon -3$$
which leads to
$$x>\sqrt{\frac 7 \varepsilon -3}$$
or also, as you teacher suggests, for any $\varepsilon>0$
$$x>\sqrt{\left|\frac 7 \varepsilon -3\right|}$$

Your way is not wrong when we deal with $x$ sufficiently large (i.e. $x>1$), such that $x<x^2$, and for any fixed $0<\varepsilon<\frac 7 4$, indeed by the definition of limit if we have
$$ x>x_0=\sqrt{\frac 7 \varepsilon-3}>1 \implies \frac{7}{x^2 +3}<\varepsilon $$
by
$$\frac{7}{x +3}<\varepsilon \implies x\ge x_1=\frac 7 \varepsilon-3>1$$
we also have that
$$x_1=\frac 7 \varepsilon-3>\sqrt{\frac 7 \varepsilon-3}=x_0$$
and therefore assuming $x>x_1$ suffices.
For $\varepsilon\ge \frac 7 4$ we just need $x>1$.
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