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

Precalculations

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

$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|$

$=|3(x+2)^2-4(x+2)|$

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

$<3\delta^2+4\delta$

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

Proof

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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    @SingaporeanDude. Thanks. Happy to help!2012-09-30

2 Answers 2

4

You made a mistake here:

$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|=|3(x+2)^2-4(x+2)|$

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:

$|3x^2+4x-2-2|=|3(x+2)^2-8(x+2)|=$

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

Thus, we're fine if

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

Thus we can choose

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

1

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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    Thanks Martin, I appreciate it. I'm not really used to using LaTeX. =]2013-05-09