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Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.


My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that $$0<|x+2|<\delta\ \ \text{implies}\ |3x^2+4x-2-2|<\epsilon$$



$\leq3|x+2|^2+4|x+2|$ by triangle inequality


Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$


For all $\epsilon>0$, choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$

$$\begin{align*}0<|x+2|<\delta\ \ \to\ \ &|3x^2+4x-2-2|<3\delta^2+4\delta\\&<3\left(\sqrt{\frac{\epsilon}{6}}\right)^2+4\delta\\&=\frac{\epsilon}{2}+4\delta\\&<\frac{\epsilon}{2}+4\frac{\epsilon}{8}\\&=\frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}$$

Therefore proven? Hehe. Not sure this will work or not. My doubts lies in the steps.

Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$

choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$

And hey, I am looking out for other possible ways to do this question too.

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Never mind: I see now what you were doing. – Brian M. Scott Sep 29 '12 at 9:12
It looks fine to me, and that seems as efficient an approach as any. – Brian M. Scott Sep 29 '12 at 9:20
There are usually many possible choices of $\delta$. By the way, you could take $\delta=\min\{1/8,\epsilon/8\}$: if $\epsilon\le 1$, then $\epsilon/8<\epsilon/6<\sqrt{\epsilon/6}$. – Brian M. Scott Sep 29 '12 at 9:28
Just a comment in order to get shorter arguments: When you reach $3\delta^2+8\delta$, which you only need to be $\leq\varepsilon$, you may argue that the $\delta$ you look for should be small, say $\delta\leq1$ and then $3\delta^2+8\delta\leq 3\delta+8\delta =11\delta$. Hence, $\delta\leq\min(1,\varepsilon/11)$ will do. – AD. Sep 29 '12 at 11:22
+1) For a well written question where you gave your thoughts! – AD. Sep 29 '12 at 11:26
up vote 3 down vote accepted

You made a mistake here:


It should be $\,8\,$ instead $\,4\,$ in the RHS. All the rest you did is fine, fixing this little mistake.

I show you now how'd I do it:


$$|x+2|\,|3x-2|\stackrel{\text{for}\,|x+2|<0.5\Longrightarrow |3x-2|<10}<10|x+2|$$

Thus, we're fine if

$$10|x+2|<\epsilon\Longrightarrow |x+2|<\frac{\epsilon}{10}$$

Thus we can choose

$$\delta =\min\left(\frac{\epsilon}{10}\,,\,\frac{1}{24}\right)$$

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Here's how I would do it:

For any $\epsilon>0$, choose $\delta=\min\left(1,\dfrac{\epsilon}{11}\right)$. Then:

$$\begin{align*} 0<|x+2|<\delta\ \ \implies\ \ |(3x^2+4x-2)-2| &= |3x^2+4x-4| \\ &= |3x-2||x+2| \\ &= 3\left|x-\frac{2}{3}\right||x+2| \\ &= 3\left|x+2-\frac{8}{3}\right||x+2| \\ &\le 3\left(|x+2|+\left|\frac{-8}{3}\right|\right)|x+2| \text{ }\text{ }\text{ by triangle ineq.}\\ &< 3\left(1 + \frac{8}{3}\right)|x+2| \text{ }\text{ }\text{ since }|x+2|<\delta\le1\\ &=11|x+2| \\ &<11\left(\frac{\epsilon}{11}\right) \text{ }\text{ }\text{ since }|x+2|<\delta\le\frac{\epsilon}{11}\\ &=\epsilon\end{align*}$$

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You can use \text command in math mode. It preserves spacing and uses normal fonts (not the fonts used in math mode). For example $x\text{ and }y$ renders as $x\text{ and }y$. – Martin Sleziak May 9 '13 at 13:04
Thanks Martin, I appreciate it. I'm not really used to using LaTeX. =] – Adriano May 9 '13 at 20:06

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