Is my proof that $ \lim\limits_{x \to 2} \left(x^2 - 3x\right) = -2 $ correct?

Question

My proof: take $x$ such that $|{x - 2}| < 1$. Since $|{x}| -|{2}| < |{x - 2}|$, then $|{x}| < 3$. Now we want to find a number $n$ such that $|{x - 2}| <1 \leq \frac{\epsilon}{n} $ or $|{x - 2}| < \frac{\epsilon}{n} \leq 1 $ and also so that $|{x - 1}| < n$, and since $|{x}| - |{1}| < |{x - 1}|$ we have that $|{x}| < n + 1$ so we can choose $n = 2$ and by definition of limit, for all $\epsilon$ we have at least one $\delta$ equal to the smaller between 1 and $\frac{\epsilon}{2}$ such that $|{x - 2}| < \delta$ implies $|{x - 1}||{x - 2}| = |{x^2 - 3x - (-2)}| < \frac{\delta}{2} = \epsilon$.

Answer 1

No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $\delta$ must exist. But the goal is to prove that such a $\delta$ exists. And how do you know that $1\leqslant\frac\varepsilon n$?

You can do it as follows: start assuming that $\lvert x-2\rvert<1$. It follows from this that $\lvert x-1\rvert<2$. Now, given $\varepsilon>0$, take $\delta=\min\left\{\frac\varepsilon2,1\right\}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $\lvert x-2\rvert<\delta$, you have\begin{align}\lvert x^2-3x-(-2)\rvert&=\bigl\lvert(x-1)(x-2)\bigr\rvert\\&=\lvert x-1\rvert.\lvert x-2\rvert\\&<2\times\frac\varepsilon2\\&=\varepsilon.\end{align}