I want to show that

$$ \lim\limits_{x\to 0^{-}} (x^2+cx+d) = d $$

Here is my attempt:
We want to show that

$$ \forall\varepsilon>0,\exists\delta>0 : -\delta\leq x\leq 0\implies\left\lvert x^2+cx + d - d\right\rvert = \left\lvert x^2 +cx\right\rvert \leq\varepsilon \tag{1}\label{1} $$

Fix an $\varepsilon>0$ and put $\delta = \min\left(1,\frac{\varepsilon}{1+\left\lvert c\right\rvert}\right)$.

By the triangle inequality we have for $x\in[-\delta,0)$ :

$$ \left\lvert x^2 +cx\right\rvert\leq \left\lvert x\right\rvert^2 +\left\lvert c\right\rvert\left\lvert x\right\rvert\leq \delta^2 +\left\lvert c\right\rvert\delta\leq\delta\left(1 + \left\lvert c\right\rvert\right) = \frac{\varepsilon(1+\lvert c\rvert)}{1+\lvert c\rvert} = \varepsilon \tag{2}\label{2} $$

The minimum here ensures us that when $\varepsilon$ is "big" $($namely $\varepsilon > 1+\left\lvert c\right\rvert)$ we still have $\eqref{2}$ which is satisfied.

EDIT : Many thanks to Surb for the correction et Nerrit for improving the format of my question.

  • 1
    $\begingroup$ The arguments are there, I would just write the final proof in a slightly different way. ... Fix an $\epsilon>0$ and set $\delta = \min(1,\epsilon/(1+|c|))$. Let $x\in [-\delta,0]$ then $|x^2+cx| \leq \ldots \leq \epsilon$ which shows that $\lim$... $\endgroup$
    – Surb
    Jan 13, 2023 at 21:04
  • $\begingroup$ @Surb Thank you for your comment ! You are right ! I need to respect the order of the definition of the limit in order to make my proof clear, I will edit this right now. $\endgroup$
    – coboy
    Jan 13, 2023 at 21:21
  • 1
    $\begingroup$ I think this looks good! By the way, in (2), after the second inequality you may want to have $\delta^2+|c|\delta$ instead. $\endgroup$
    – Surb
    Jan 13, 2023 at 22:58
  • 1
    $\begingroup$ Yes you are right ! Thank you a lot really ! $\endgroup$
    – coboy
    Jan 13, 2023 at 23:03


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