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I have to prove the following using epsilon delta: $$\lim\limits_{x \to -2} x^3+4x^2+4x-1 = -1$$

if $|x+2| < \delta$ then $|f(x) - f(a)| < \varepsilon$

Factorizing f(x): $|(x+2)(x^2+2x)| < \varepsilon$

Divide both sides by $|x^2+2x|$ and get: $|x+2| < \frac{\varepsilon}{|x^2+2x|}$.

I have $-\delta < x+2 < \delta$ from $|x+2| < \delta$

I then set $\delta:=1$ and subtract $2$ from all sides: $-3 < x < -1$

I do this to create an upper bound for $x$ and therefore an upper bound for $|x^2+2x|$

How do I know which side of the inequality (-3 or -1) to substitute for $x$ in $\frac{\varepsilon}{|x^2+2x|}$?

Alternatively, if thats the wrong approach, what's the right approach for deciding $\delta$?

eh98
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    Welcome to StackExchange. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – md2perpe Aug 30 '20 at 19:30
  • Why do you want to use one of $-3$ and $-1$ in $|x^2+2x|$? – md2perpe Aug 30 '20 at 19:34
  • It's unclear what you're doing or why you think it would be any use. I suggest you show all the steps with explanation. – David K Aug 30 '20 at 20:26
  • Sorry, i tried to clean it up a bit now, I want to use one of $-3$ and $-1$ to create an upper bound for $|x^2+2x|$ such that $|x+2| |x^2+2x| < \delta * upperBound$ – eh98 Aug 31 '20 at 11:25

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For any $x$ such that $|x+2|<1$ we have that $|x|<3$, and so, for all $\varepsilon>0$, if $x$ satisfies that $$|x+2|<\min\Big\{1,\sqrt \frac{\varepsilon}{3} \Big\}$$ then $\displaystyle |(x^3+4x^2+4x-1)-(-1)| = |x(x+2)^2| = |x||x+2|^2 < 3 \Big( \sqrt \frac{\varepsilon}{3} \Big)^2 = \varepsilon.$

azif00
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  • Could you elaborate on your scratchwork from $|x||x+2|^2$? Thank you! – eh98 Aug 30 '20 at 22:48
  • You mean the last inequality? – azif00 Aug 30 '20 at 23:30
  • Yes, how do you get $\sqrt{\varepsilon/3}$? The more I look at it, the more I think $\delta$ should be minimum of $1$ and $\varepsilon/3$ – eh98 Aug 31 '20 at 11:17
  • If $|x+2| < \min{1,\varepsilon/3}$, $|x| < 3$ and $|x+2|^2 < \varepsilon^2/9$, meaning that $|x||x+2|^2 < \varepsilon^2/3$. – azif00 Aug 31 '20 at 19:06