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Find the limit $L$ using the definition of limit. Then find $\delta>0$ such that $|f(x)-L|<0.01$ whenever $0<|x-c|<\delta$. $$\lim_{x\to 2}(x^2-3)$$

I have gotten as far as $|x-2|=0.01/|x+2|$, and I need help with further steps, I do not know what more there is to do to find delta.

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Andrés E. Caicedo
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user170112
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    what have you tried so far? give that in as much detail please..even if it is the other questions and then show where you get stuck on the question in particular. If writing equations is a problem use this tutorial. with "\epsilon" and "\delta" coming in handy here. Thanks! – Chinny84 Aug 16 '14 at 23:40
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    I have gotten as far as |x-2|=0.01/|x+2|, and I need help with further steps, I do not know what more there is to do to find delta. – user170112 Aug 16 '14 at 23:47
  • Here is a similar problem, although $x\to1$ in that case: http://math.stackexchange.com/questions/209440/how-to-show-that-fx-x2-is-continuous-at-x-1 – Martin Sleziak Aug 17 '14 at 07:31
  • It seems that the OP have included their thoughts in a comment. They were edited by another user into the post. For these reasons I voted to reopen. – Martin Sleziak Aug 18 '14 at 06:39
  • @user170112: Check out this. – Mhenni Benghorbal Aug 24 '14 at 04:30

2 Answers2

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If you choose $$\delta:=\min\{1,\frac\varepsilon5\}$$ then for $|x-2|<\delta$ you have $|x-2|<1$, which implies $1<x<3$ and $|x+2|<5$.

So you get $$|x^2-2|=|x+2|\cdot|x-2|<\frac\varepsilon5\cdot 5 = \varepsilon.$$

Martin Sleziak
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  • I have understood your solution as to proving the epsilon which is 0.01, but are the intervals known because of 2, 1 and 3 are the outside number or could it have been 1.9 and 3.1, or any numbers surrounding 2? – user170112 Aug 18 '14 at 13:34
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$$\lim_{x \rightarrow 2} (x^2-3)=2^2-3=4-3=1$$

The limit $L$ is equal to $1$.

We have to find a $\delta>0$ such that $|f(x)-1|<0.01$ whenever $0<|x-2|<\delta$.

EDIT: $$|x-2|<\delta \Rightarrow -\delta<x-2<\delta \Rightarrow 2-\delta<x<2+\delta$$

$$|f(x)-1|<0.01 \Rightarrow |x^2-3-1|<0.01 \Rightarrow |x^2-4|<0.01 \Rightarrow -0.01<x^2-4<0.01 \\ \Rightarrow 3.99<x^2<4.01 \Rightarrow \sqrt{3.99}<x<\sqrt{4.01}$$

Therefore, we could set $2+\delta=\sqrt{4.01} \Rightarrow \delta \approx 0.002$

Mary Star
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  • Thank you, But my book says delta=0.002, I do not know as to how they have gotten to this conclusion. – user170112 Aug 16 '14 at 23:56
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    $\delta$ should depend on the $\underline{\text{fixed}}$ $\varepsilon$ given, not $x$. $x$ is a $\underline{\text{variable}}$ that could be anywhere within some range of the number 2. – mjh Aug 17 '14 at 03:36
  • Can you explain a bit more about the range you are speaking of mjh? – user170112 Aug 17 '14 at 04:42
  • By range I mean the interval $(c-\delta,c+\delta)$. The crux of the problem is that you are given an arbitrary $\varepsilon$ and you need to find a $\delta$ so that $x$ being in the interval $(c-\delta,c+\delta)$ will assure us that $f(x)$ falls into the interval $(f(c)-\varepsilon,f(c)+\varepsilon)$. – mjh Aug 17 '14 at 04:52
  • Sorry, but how would I know the interval? – user170112 Aug 17 '14 at 05:06
  • That's the thing, you need to find the interval. Martin Sleziak gives a good answer above. He let $\delta=\min{1,\varepsilon/4}$ so his interval is either $(c-1,c+1)$ or $(c-\varepsilon/4,c+\varepsilon/4)$, depending on how small the $\varepsilon$ given is. – mjh Aug 17 '14 at 05:11
  • I edited my answer! – Mary Star Aug 18 '14 at 00:24
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    I understood how you have done this Mary, the solution is showing a different form of your answer 0.01/5=0.002 http://c86f903f96a3ec2dd206-646e12b456b1588b6ef5c188c33a829b.r15.cf2.rackcdn.com/se01b01035.png – user170112 Aug 18 '14 at 13:10