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Let we have

\begin{equation*} n\times\frac{1}{n}=\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}\mbox{ ($n$ times)}. \end{equation*}

Taking $\lim_{n\to\infty}$ to both sides, we get

\begin{eqnarray*} \lim_{n\to\infty}1 &=& \lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}\right)\\ \Longrightarrow1&=&0+0+\cdots+0\mbox{ ($n$ times)}\\ &=&0. \end{eqnarray*}

I am not an expert of math, and confused that where the confusion is.

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    $\begingroup$ An informal answer here: The problem is the "dot...dot...dot" part. There is such thing as an indeterminate form, being zero times infinity. And that is typically NOT zero. With the "dot...dot...dot" notation, you are "reading" the problem as a finite amount of $1/n$ terms (even though it is not your intention) and hence you come up with zero as the (incorrect) answer. $\endgroup$ – imranfat Aug 5 at 2:17
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    $\begingroup$ Also, I up-voted the question because it is a good question. You pose this question to introductory calculus students and I bet half of them cannot point out the issue. $\endgroup$ – imranfat Aug 5 at 2:19
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    $\begingroup$ By your logic, on the left you also have $\lim n \cdot \frac{1}{n} = \lim_{n\to\infty} n \cdot \lim_{n\to\infty} \frac{1}{n} = \infty \cdot 0.$ $\endgroup$ – mjw Aug 5 at 2:20
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    $\begingroup$ The problem is that you can't only take limit on $n$ only on part of the expression. There is the "$n$ times" which also depended on the $n$ you are taking to $\infty$. And no, you can't split the limits up in two in general. $\endgroup$ – user10354138 Aug 5 at 2:21
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    $\begingroup$ See math.stackexchange.com/questions/2687406/… $\endgroup$ – Angina Seng Aug 5 at 2:27
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This is an excellent argument that we cannot in general find a limit by taking the limits of the parts of an expression.

When many students are first introduced to limit laws, they see their instructor go through a lot of complicated math in order to prove things that feel obvious. In this case, the relevant one is the addition law:

$$\lim_{x \to c}\left(f(x) + g(x)\right) = \lim_{x\to c}f(x) + \lim_{x\to c}g(x)$$

This seems obvious, right? A limit means "what number does this expression get close to". Of course $f(x) + g(x)$ would get close to the sum of whatever $f(x)$ gets close to and whatever $g(x)$ gets close to. So why does the instructor (or the textbook) spend half a page messing around with $\epsilon$s and $\delta$s to prove the law?

The answer is because of exactly the sort of thing you've pointed out. There are situations where the "intuitive" approach to limits stops working, essentially because infinity is hard. For those situations, we need to rely on the proof. Crucially, in this case, the proof relies on there being only two things added together. This means, if we want to adhere perfectly to the law as stated, we have to jump through hoops like this:

\begin{align*} \lim_{x \to c} \left(f(x) + g(x) + h(x)\right) &= \lim_{x \to c} \left(\left(f(x) + g(x)\right) + h(x)\right)\\ &= \lim_{x \to c}\left(f(x) + g(x)\right) + \lim_{x \to c}h(x)\\ &= \lim_{x \to c}f(x) + \lim_{x \to c}g(x) + \lim_{x \to c}h(x) \end{align*}

We can do the same to deal with four, or five, or five hundred things added together. But how would we deal with $n$ things added together, when $n$ changes over the course of the limit? If we "peel off" one like I did above, there'd still be infinitely many left over. In other words, even with aggressive uses of this limit law, we can only handle sums of fixed size. One that "grows", like $\frac1n + \frac1n + \cdots + \frac1n$ does, can't be handled this way.

To summarize: Many of the limit laws feel like they're just saying "take the limit of the parts of the expression". This isn't true; in fact, they're saying "here is one precise way in which you can find a limit by using the limits of the parts". If you want to do something to a limit that isn't one of the standard limit laws, you're doing something special, which means you'll need to go back to the definition of the limit (or something similar) in order to make sure that what you're doing works.

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  • $\begingroup$ Excellent answer. I don't know if this is worth noting, but we could peel off one term (or finitely many) from the expression in question - we just are always left with a sum that doesn't have a fixed size. e.g. $\displaystyle{\lim_{n\to\infty}}\underbrace{\frac1n+\frac1n+\cdots}_{n\text{ terms}}=\displaystyle{\lim_{n\to\infty}}\frac1n+\displaystyle{\lim_{n\to\infty}}\underbrace{\frac1n+\frac1n+\cdots}_{n-1\text{ terms}}$ $\endgroup$ – Mark S. Aug 5 at 13:05

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