$\lim _{ n\rightarrow \infty }{ { \left( -1+\frac { 1 }{ n } \right) }^{ n } } =? $

Question

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?

Answer 1

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 $k$.

Answer 2

We have that

$$\left(-1+\frac1n\right)^n=(-1)^n\left(1-\frac1n\right)^n$$

so the limit cannot exist since

$$\left(1-\frac1n\right)^n\xrightarrow[n\to\infty]{}e^{-1}>0$$

Answer 3

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

Answer 4

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

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