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I have been given this problem: For the limit $$\lim_{x\to 2}({x^3-3x+4})=6$$

illustrate "Definition 2" (I have included this below) by finding values of $\delta$ that correspond to $\varepsilon=0.2$ and $\varepsilon=0.1$

"Definition 2:" My textbook says to this definition "Let $f$ be a function defined on some open interval that contains the number $a$, except possibly $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim_{x\to a}f(x)=L$$ if for every number $\varepsilon>0$ such that $$if\;0<|x-a|<\delta\;\;\;then\;|f(x)-L|<\varepsilon$$

This has been a seriously frustrating problem, I've been working on it for two days, I just don't understand this concept at all. The only thing I think I understand is that the values of $\delta$ correspond to values on the x-axis, and $\varepsilon$ is related to the limit.

How do I go about solving this problem?

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    $\begingroup$ Did you mean limit as $x\to 0$? $\endgroup$
    – Zach Stone
    Jun 18, 2015 at 2:30
  • $\begingroup$ @ZachStone I did mean 2, but I put the wrong equation up! I was looking at the page and mixed up the equation right underneath this one that I'm working on. Thank you for pointing that out. $\endgroup$
    – matryoshka
    Jun 18, 2015 at 2:33
  • $\begingroup$ Ah, i see. This problem is somewhat more approachable $\endgroup$
    – Zach Stone
    Jun 18, 2015 at 2:34

2 Answers 2

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I think you're close to getting it to "click". What you said at the end is correct: the $\delta$ value represents the distance between $x$ and $a$, and the $\epsilon$ value represents the distance between $f(x)$ and the limit $L$.

So the $\delta$, $\epsilon$ limit definition is a quantification of the statement "As $x$ becomes closer to $a$, $f(x)$ becomes closer to $L$".

For your problem at hand, you are given $\epsilon = 0.1$. So what they're asking is "How close does $x$ have to be to $a = 2$ before $f(x) = x^3 - 3x + 4$ is at most $0.1$ away from $L = 6$?"

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  • $\begingroup$ Can I think of it like $\lim_{x\to a}x^3-3x+4=6.1$? Or $\lim_{x\to a}x^3-3x+4=5.9$? Am I trying trying to find a new value for $a$? $\endgroup$
    – matryoshka
    Jun 18, 2015 at 2:51
  • $\begingroup$ @Grace In order to understand this concept you have to work with a specific value for $a$. $\endgroup$
    – user42912
    Jun 18, 2015 at 4:26
  • $\begingroup$ @Grace, not equals, but "near". And no, you're not finding a new value of $a$, you're finding the distance $\delta$. So you want to find how close $x$ needs to be to $a = 2$ so that $|x^3 - 3x + 4 - 6| \le 0.1$. $\endgroup$ Jun 18, 2015 at 5:57
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These kinds of problems are often done backwards. We start with what we want. $$ |x^3-3x+4 -6|<\varepsilon $$ for $x$ near $2$. So replace $x$ with $2+\delta$ for some $|\delta|$ small.

Now, we can start simplifying the left. $$ |(2+\delta)^3-3(2+\delta)-2|=|8+4\delta+2\delta^2+\delta^3-6+\delta-2|=|\delta^3+2\delta^2+5\delta| $$

So what $\delta$ makes $|\delta^3+2\delta^2+5\delta|<\varepsilon$? Factoring $\delta$ gives $$ |\delta|\;|\delta^2+2\delta+5| $$

If $\delta$ is already less than 1/3, then the right factor is less than 6. So we have we want to choose $\delta$ so that $\delta 6<\varepsilon$

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