We know that $$\lim _{ n\rightarrow \infty }{ { \left( 1-\frac { 1 }{ n } \right) }^{ n } } =\frac { 1 }{ e } .$$ However the result of $$\lim _{ n\rightarrow \infty }{ { \left( -1+\frac { 1 }{ n } \right) }^{ n } } $$ is shown in complex form by Wolframalpha . Why complex numbers?

Yes, $-1+\frac { 1 }{ n } < 0 $, but if we write the values from $n=1,2..,10$ , all values will be real. Any opinion? How do you calculate this limit?


We have that


so the limit cannot exist since


  • $\begingroup$ So if the limit does not exist why Wolfram says $\\ \lim _{ x\rightarrow \infty }{ { \left( -1+1/x \right) }^{ x } } ={ e }^{ \left( 2i0\quad to\quad \pi \right) }$? $\endgroup$ – newzad Nov 6 '13 at 5:34
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    $\begingroup$ As in many other cases, Wolfram must be high...Don't trust too much machines. $\endgroup$ – DonAntonio Nov 6 '13 at 5:35

Wolfram is interpreting $x$ as a real variable, not an integer one. Hence the value is complex for some values of $x$ - in particular, $k + 1/2$ will give a complex result for every integer $k$.


$$ \lim_{n\to\infty} \left(-1+\frac{1}{n}\right)^n= \lim_{n\to\infty} (-1)^n\left(1-\frac{1}{n}\right)^n =\lim_{n\to\infty} \frac{(-1)^n}{e}=\lim_{n\to\infty} e^{n\pi i-1}$$ As a periodic function without a decrease in magnitude, there is no well-defined limit.

  • $\begingroup$ Your second equality is highly doubtable and anyway it'd require a proof: you're taking the limit of part of the sequence while still leaving other part of it untouched... $\endgroup$ – DonAntonio Nov 6 '13 at 5:38
  • $\begingroup$ Yes. $\lim_{n\to a} f(n)g(n)=\left(\lim_{n\to a} f(n)\right)\left(\lim_{n\to a} g(n)\right)$. See: Theorem and Proof 3 tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx $\endgroup$ – Tim Ratigan Nov 6 '13 at 5:52
  • $\begingroup$ @DonAntonio you can see the last equality by writing (-1) as $e^{n\pi i}$. $\endgroup$ – Betty Mock Nov 6 '13 at 7:10
  • $\begingroup$ @Tim.Ratigan -- it's not necessary to change it to complex form. The sequence {1/e, -1/e, 1/e, -1/e ...} clearly cannot converge to anything. $\endgroup$ – Betty Mock Nov 6 '13 at 7:16
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    $\begingroup$ @Tim.Ratigan, the equality $\;\lim f(n)g(n)=\lim f(n)\lim g(n)\;$ can be assured only if both limits exist separatedly and this is not the case. $\endgroup$ – DonAntonio Nov 6 '13 at 11:00

To understand the answer, you need to know three things:

  • Wolfram is taking your limit over the reals
  • Wolfram interprets exponentiation as complex exponentiation (and probably takes the principal value, but that's not relevant here)
  • There is a useful notion called a "limit point"

A limit point is much like a limit. If $L$ is the limit of $f(x)$ as $x$ approaches $a$, that means we need $f(x)$ to be near $L$ for all $x$ near $a$.

However, if we merely require that this be true for infinitely many $x$ near $a$ (i.e. for a sequence of $x$'s that converge to $a$), we instead get the notion of "limit point".

A good example of this notion is the limiting behavior of $\sin(x)$ as $x \to +\infty$: the set of limit points for this is the entire interval $[-1, 1]$, since $\sin(x)$ keeps oscillating continuously from $-1$ to $1$ as $x$ grows. Wolfram agrees.

An simplistic example of something useful you can do with limit points is to compute:

$$ \lim_{x \to +\infty} \frac{1}{x} \sin(x) = \lim_{x \to +\infty} \frac{1}{x} \cdot \lim_{x \to +\infty} \sin(x) = 0 \cdot [-1,1] = 0$$

That said, I think Wolfram still made an error in its simplification: the set of limit points for your limit, I believe, should have been listed as

$$ e^{-1 + 2 \mathbf{i} x} \qquad \qquad x \in [0, \pi] $$

  • $\begingroup$ Actually, despite my last comment, I expect that wolframalpha did something correct based based on a reasonable interpretation of the problem -- it's just not clear what precise problem it decided to solve. $\endgroup$ – user14972 Nov 6 '13 at 5:50
  • $\begingroup$ Shouldn't it be $e^{-1 + n\pi i}$? $\endgroup$ – Betty Mock Nov 6 '13 at 7:18
  • $\begingroup$ @Betty: I was taking the limit over the reals rather than the naturals. If you restrict to the naturals, then yes: or more simply, the set of limit points would be $\pm e^{-1}$, in harmony with the fact that the limit points to $(-1)^n$ are $\pm 1$. $\endgroup$ – user14972 Nov 6 '13 at 7:33
  • $\begingroup$ @Hurkyl , I think you're right and WA gives the set of limit points of a sequence, but I think it is a serious offence (?) to do so without explicitly saying so. $\endgroup$ – DonAntonio Nov 6 '13 at 11:24
  • $\begingroup$ @DonAntonio I agree it is an offense. But on the side of patience I would say that whenever anyone writes a lot of material many things will not be pristine. It is also hard when you know a very great deal to understand what is confusing to those with less knowledge. $\endgroup$ – Betty Mock Nov 7 '13 at 1:35

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