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I don't quite understand what it really means. Do it mean that $f(c_i)$ could be infinite and $dx$ is very small, so you can't determine what the infinite$\times$very small is?

Berkheimer
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whoisit
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2 Answers2

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The point is that if the function is nice, then the difference between all the different $f(c_i)$ for all valid choices of $c_i$ are roughly the same. Therefore, for any given partition, if it is fine enough, changing $c_i$ to some other valid value makes very little change in the total sum.

However, if $f$ is unbounded, then you can choose $c_i$ that makes $f(c_i)$ as large as you want, which means that for any given partition, different choices of $c_i$ can make that sum be basically whatever you want. That is why the limit is not defined.

Arthur
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  • But even though $f(c_i)$ coule be as large as I want, but it's still multiplied by very small difference of partition, I still have on control on making it whatever I want. – whoisit Feb 17 '16 at 23:29
  • @whosit Not really. I am, for a second, ignoring the limit symbol and just looking at one specific partition. I have $\Delta x_i$ available, and it does not change. Therefore I can make not only $f(c_i)$ as big as I want by choosing $c_i$, I can make $f(c_i)\Delta x_i$ as big as I want. This means I can make the sum over that specific partition as big as I want. Only after that do I look at the $\lim$ operator and conclude that it doesn't make sense. – Arthur Feb 18 '16 at 07:37
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Take a specific example, say $\int_0^2\frac 1{(x-1)^2}dx$ One of the intervals will cover $1$. The definition says the sum must converge no matter how we choose the $c_i$ in that interval (and all the rest), but if we choose that $c$ to be $1$ the sum is not defined. Even if we do not choose exactly $1$, we can choose it close enough to $1$ that the sum is enormous. As the text says, that shows we have to do something more carefully. Presumably the next paragraphs will show what that is.

Ross Millikan
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  • Thank you for your answer. I still don't get it. I can understand what we choose the c1 to be 1, which is not defined. But for $$\lim_{n\to \infty}f(c_i)\Delta x,$$ $f(c_i)$ is $\infty$ and $\Delta x$ is very small, what could it be when $\infty$*very small? it's hard to say by choosing c1, we could get the sum as large as we want – whoisit Feb 18 '16 at 00:11
  • It comes down to the order of choosing the points. In the basic definition, $\Delta x$ is chosen first, so $c$ can be chosen to make $f(c)$ larger than $\frac 1{\Delta x^2}$ so the product is large. The limit your book will show you next puts a cap on how big $f(c)$ can be, then lets you choose $\Delta x$ to keep the product small. – Ross Millikan Feb 18 '16 at 00:48
  • $f(c_i)$ is large ($\infty$), $\frac{1}{\Delta x^{2}}$ is also large($\infty$), how can we tell which one is larger? It is a little bit hard to choose a $\infty$ larger than another $\infty$ The book didn't show how big $f(c_i)$ can be, it just gave an example. – whoisit Feb 18 '16 at 01:23
  • I was choosing $c_i$ so $f(c_i)=\frac 1{\Delta x^2}$ so the product with $\Delta x$ is large. – Ross Millikan Feb 18 '16 at 01:35
  • em, maybe I got sth wrong in my mind. could you please tell me what's the limit of $$\lim_{x\rightarrow\infty ,y\to 0}xy$$ – whoisit Feb 18 '16 at 02:59
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    There is no limit to this because you let both variables go simultaneously. But $\lim_{x \to \infty}(\lim_{y \to 0} xy)=0$ because you do the $y$ limit first. – Ross Millikan Feb 18 '16 at 04:11
  • the product is large, we can say the intergal must diverge(it should be divergent, but not always the case actually) – whoisit Feb 22 '16 at 07:35