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
 A: Your solution is almost ok.
Indeed, you need to show that the inequality $\varepsilon>7/(x^2+3)$ holds for all sufficiently large $x$. In the next step you can simplify the resulting inequality by assuming that $x>1$ (surely ok, when $x\to\infty$). This allows you to estimate:
$$
\frac7{x^2+3}<\frac7{x+3}
$$
as decreasing the denominator increases a positive fraction. As we have assumed that $x>1$, we also have $x+3>0$, whence the inequality $7/(x+3)<\varepsilon$ is true whenever $x>7/\varepsilon-3$, and the desired inequality is a consequence of this.
Then you can wrap up, and say that
$$\left|\frac{1-2x^2}{x^2+3}-(-2)\right|<\varepsilon$$
for all $x>\max\{1,7/\varepsilon-3\}$.

Forgetting that $\max$ and comparison with $1$ is not much of problem. As a teacher I might, ideally, wish you to explain the logic more carefully, but you are still learning how to do it. So I think overall you did fine. Don't know what your teacher is thinking in their complaint. When proving a limit you hardly ever need to find the tightest possible bound, when a looser one gets the job done.

I would write your second line (with the caveat about $x>1$ in place) as follows: If $7/(x+3)<\varepsilon$, then also $7/(x^2+3)<\varepsilon$. This would make the logic clearer. This kind of proofs are often reverse-engineered. And if you have only seen reverse-engineered proofs, abbreviated on the chalkboard with verbal explanations only, copy/pasting whatever is on the chalkboard may look funny. As is the case here. When you write
$$\varepsilon>\frac7{x+3}>\frac7{x^2+3}$$
it is not entirely clear what is meant. Was this supposed to be a consequence of the preceding line? Surely not! After reverse-engineering it would be clear. If I did this on a chalkboard, I would, like you, first jot down the target inequality $7/(x^2+3)<\varepsilon$. Then I would proceed with the simplifying estimate $7/(x^2+3)<7/(x+3)$.... and then ... I might have left empty space at the exact spot where you wrote:
$\varepsilon>7/(x+3)>7/(x^2+3)$, and write the same thing :-).

In my opinion any complaint at your solution, as written in the question box, should be about the lack of explanations.

A: What makes you think that $\varepsilon > \frac{7}{x + 3}$?
In fact, given only that
$\left\lvert\frac{-7}{x^2 +3}\right\rvert < \varepsilon,$
it is possible to choose $x$ such that
$$ \frac{7}{x^2 + 3} < \varepsilon < \frac{7}{x + 3}.$$
It might be instructive for you to try to find such an $x$ for some particular value of $\varepsilon,$ let's say $\varepsilon = 0.001.$

That is likely what your teacher was thinking.
However, there is another viewpoint.
We are trying to find an $N$ such that if $x > N$ then
$\left\lvert\frac{-7}{x^2 +3}\right\rvert < \varepsilon.$
We can show that if
$N = \sqrt{\left\lvert\frac7\varepsilon - 3\right\rvert}$
then this value of $N$ is sufficient.
But if that value of $N$ is sufficient, then any larger value of $N$
is also sufficient.
And the proof only requires us to find a sufficient value of $N$,
not the smallest sufficient value of $N$.
In other words, it is good enough to choose any $N$ such that
$N \geq \sqrt{\left\lvert\frac7\varepsilon - 3\right\rvert}.$
Provided that $\frac7\varepsilon - 3 > 1,$
we have
$\frac7\varepsilon - 3 > \sqrt{\left\lvert\frac7\varepsilon - 3\right\rvert}.$
So if you set $N = \frac7\varepsilon - 3$
you have $N \geq \sqrt{\left\lvert\frac7\varepsilon - 3\right\rvert}$
and you should be able to finish the proof.
But there is a caveat here, which is that for some (large) values of $\varepsilon$ we have $\frac7\varepsilon - 3 < 1.$
By setting $N = \frac7\varepsilon - 3$ in those cases you get an $N$
that is not large enough; it might even be negative.
So you have to say something about those cases, such as setting
$N = 1$ in case $\frac7\varepsilon - 3 < 1.$

What I think you should learn from this is not that one method is better than the other. Rather, the lesson is that regardless of which method you use,
it is important to explain why you are taking each step so that the reader can see that the step is a correct step for that purpose.
Indeed, strictly speaking, the calculations you showed are not even part of the proof. Yes, you must somehow devise a way of setting $N$ such that
$x > N$ implies $\left\lvert\frac{-7}{x^2 +3}\right\rvert < \varepsilon,$
but you don't actually own the reader an explanation of how you came up with that particular formula for $N.$ All you need to show is that $N$ produced in this particular way will always work.
But if you do show these calculations, you need to explain them.
A: 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 $x>1$.
