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 $(x+1)(2x+2)$.
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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 Apr 30 '12 at 04:40
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6You 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 05:06
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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. – davidlowryduda Apr 30 '12 at 05:12
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1It would be useful to know why that doesn't match your definition. – Pedro Apr 30 '12 at 05:31
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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 07:35
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A computation mistake might be at the root of this question. Let $u(x)=2x^2+5x+3$. To show that $\lim\limits_{x\to-2}u(x)=1$, one can look at $u(x)-1=(x+2)(2x+1)$ (note the difference between this and $(x+2)(2x+2)$ written in the question). The rest is standard: assume $|x+2|\leqslant\delta$ with $\delta\leqslant1$, then $|u(x)-1|\leqslant5\delta$, hence... – Did Apr 30 '12 at 07:36
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Hello André Nicolas you are right. This is the right question. – Vinicius L. Beserra Apr 30 '12 at 14:55
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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. – Vinicius L. Beserra Apr 30 '12 at 15:05
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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.
Arturo Magidin
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@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 01 '12 at 16:14
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Ok, but what´s is the criteria to choose these values for the inequality? From Where i take it. May you answer me. I know that every question has different numbers. For example in another question as in limit x->3 x^2=9 I have to choose anoter values for the inequality. Thanks. – Vinicius L. Beserra May 09 '12 at 19:28
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@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 09 '12 at 19:48
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@Vinicius: If you are asking how I got $-3\lt x\lt -1$, it follows from assuming $|x+2|\lt 1$. which we can do because if any $\delta$ works, then so does any smaller one, so we can always assume that the $\delta$ we will pick is smaller than any particular constant we care to specify in advance (in this case, than $1$). – Arturo Magidin May 09 '12 at 19:59
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