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In Chapter 1: Functions and limits, 1.7 The Precise Definition of a Limit,

Let $f$ be a function ... 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 $\epsilon>0$ there is a number $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$.

OK, let's assume this is the case and that this definition is perfectly true, but:

  • Then how do you go about determining this $L$ by the definition? For example if I wanted to compute $\lim_{x\to0}\frac{\sin x}{x}$ how can I determine this $L$ from the definition (without pre-knowledge of the limit)? (this is related to the question below)
  • Can we find a suitable $\delta_1$ such that $|f(x)-L_1|<\epsilon$ and a different $\delta_2$ such that $|f(x)-L_2|<\epsilon$? If so then what is right $L_1$ or $L_2$?

I think if those issues are cleared to me then the rest will also be clear.

Thanks in advance.

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    $\begingroup$ And of course he can't prove his definition is correct. It's a definition. Hand-waving is always used to justify definitions. $\endgroup$ – Thomas Andrews Sep 23 '14 at 19:36
  • $\begingroup$ I'm more talking about why we choose this particular definition given those issues? Or how are those issues resolved to support the definition? $\endgroup$ – user178276 Sep 23 '14 at 19:38
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    $\begingroup$ Read on, and you will see how to compute $L$ in special cases. As noted, this is the definition, and it will give you the results you expect when you expect them. For your second issue, when $\epsilon$ beomes arbitrarily small, there can only be at most one limit $L$. Indeed, supposing there are two means $|L_1-L_2|<\epsilon$ for all $\epsilon$, so $L_1=L_2$ $\endgroup$ – Shakespeare Sep 23 '14 at 19:39
  • $\begingroup$ @user178276 I know this is old but I wanted to address your question "about why we choose this particular definition". As mathematicians, we can all define things in whatever way we want. I could make my own definitions for anything! The biggest thing, though, is finding meaning in these definitions. For example, I could define something to be the number of letters used in an equation, but the question is: "so what?" How does that help in the world of math? As it turns out, our definition of a limit helps build the foundation of calculus, which can be used to solve countless problems. $\endgroup$ – Sultan of Swing Dec 11 '14 at 12:32
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The definition doesn't show you how to determine $L$. There is no general way to determine $L$. This definition only shows that at most one $L$ exists (see below.) If there was a general way to determine $L$, calculus would be a nothing-burger.

You can prove from this definition that there is only one limit, at most.

For any particular $\epsilon$ you can find two $L_1$ and $L_2$ and $\delta_1$ and $\delta_2$, but there is no way for it to work for all $\epsilon.$ The definition says that for all $\epsilon>0$. This is provable from the definition:

If $L_1\neq L_2$ let $\epsilon=|L_1-L_2|/2$. Try to pick $\delta_1$ and $\delta_2$ for this $\epsilon$. One can prove that this is not possible.

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  • $\begingroup$ OK, show me the proof then if you are right. $\endgroup$ – user178276 Sep 23 '14 at 19:46
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    $\begingroup$ @user178276: If there are two limits, $L_1$ and $L_2$, then let $0 < \epsilon < |L_1 - L_2|/2$. Pick your $\delta_1$ and $\delta_2$. Then if $|x - a| < \min(\delta_1,\delta_2)$, we have $f(x)$ is closer than $\epsilon$ to both $L_1$ and $L_2$, which is impossible. $\endgroup$ – Nick Matteo Sep 23 '14 at 19:54
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    $\begingroup$ @user178276 That's a bit demanding. It is proven in every calculus book that is remotely rigorous. $\endgroup$ – Thomas Andrews Sep 23 '14 at 20:10
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    $\begingroup$ So, the history is that Newton used a notion of infinitesimals to define certain types of limits. This bothered quite a few people, because they didn't know if infinitesimals were "real." Lots of years went along before this definition came up, and it is irritatingly complicated, but it really does codify the intuitive notion quite well. It is frustrating at first for quite a few people, though. $\endgroup$ – Thomas Andrews Sep 23 '14 at 20:12

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