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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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    André Nicolas the question is the 13th from Louis Leithold. I will upload a imagem but i can´t do it, 'cause i don´t have permission.2012-04-30

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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.

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    Finaly i understand.2012-05-12