# Use the epsilon-delta definition to prove the following statement.

I'm trying to use the epsilon delta definition to prove that $$\lim _{x\to-2} (2x^2+5x+3)=1$$ when I have $\epsilon < 0,04$. So, I have a problem because the quadratic equation becomes $(x+1)(2x+2)$.

This is the answer based in the epsilon-delta definition of limit.

I will upload a jpg scanned image.

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Do you know the theorem that if $\lim f = A$ and $\lim g = B$ then $$\lim f\cdot g = A\cdot B?$$ – Pedro Tamaroff Apr 30 '12 at 4:40
You have given a quite incomplete version of the question. I can guess that you were asked to show that $\lim_{x\to -2}(2x^2+5x+3)=1$. I can also guess that for $\epsilon=0.04$ you were asked to come up with a $\delta$ that had a certain property. But one should not have to guess. Can you edit and give the actual question? – André Nicolas Apr 30 '12 at 5:06
I have edited the current content so that it looks nice. Please check to make sure I haven't changed anything important, and also please edit so that the question is complete. – mixedmath Apr 30 '12 at 5:12
It would be useful to know why that doesn't match your definition. – Pedro Tamaroff Apr 30 '12 at 5:31
Your problem is to show that the limit is equal to 1, so you have to work with the absolute value of the polynomial after substarcting 1, and then you can factorize and one of the factors will be X+2. – alpha.Debi Apr 30 '12 at 7:35

We want to show that $$\lim_{x\to -2}(2x^2+5x+3) = 1$$ using $\epsilon$-$\delta$. Or rather, you want to find a $\delta$ such that if $0\lt |x-(-2)|\lt\delta$, then $|(2x^2+5x+3)-1|\lt \epsilon$ for $\epsilon=0.04$.
Note that $2x^2+5x+3-1 = 2x^2+5x+2 = (x+2)(2x+1)$. So we want to control both $|x+2|=|x-(-2)|$ and $|2x+1|$. Note that if $|x+2|\lt 1$, then $-3\lt x\lt -1$, so $-6\lt 2x\lt -2$, and $-5\lt 2x+1\lt -1$, so $1\lt |2x+1|\lt 5$.
So we would like $|x+2|$ to be both less than $1$, and also less than $(0.04)/5 = 0.008$. For example, take $\delta=0.005$. If $0\lt |x+2|\lt 0.005$, then $|2x+1|\lt 5$, and we have: $$|(2x^2+5x+3)-1| = |(x+2)(2x+1)| = |x+2|\,|2x+1|\lt (0.005)5 = 0.025\lt 0.05=\epsilon$$ so this $\delta$ suffices.
@ViniciusL.Beserra: Because we have an estimate for $x$, but want an estimate for $|2x+1|$ (which is the extra factor that will appear when evaluating and factoring $|(2x^2+5x+3)-1|$. So go from an estimate to $x$ to an estimate for $2x$, to an estimate for $2x+1$, to an estimate for $|2x+1|$. – Arturo Magidin May 1 '12 at 16:14
@Vinicius: I don't understand your question. If you were doing $\lim_{x\to 3}x^2$, you would have to estimate $|x^2-9|=|x-3||x+3|$, and since you have control over $|x-3|$, you need to estimate $|x+3|$. Since we only care about $x$s that are close to $3$, and this function does not have any "weird stuff" that we need to avoid, we may assume that we will take a $\delta$ that is no larger than $1$, so that $2\lt x\lt 4$ (since we will have $|x-3|\lt 1$), and go from there. What criteria for saying "no larger than $1$"? Nothing beyond "$1$ is easy and there's nothing we need to avoid." – Arturo Magidin May 9 '12 at 19:48