Why is the $\lim_{x \to \infty} \frac{(-1)^x}{x}= 0$?

Question

I know that the answer is $0$ by using Wolfram. But I'd appreciate it if someone shows the solution to this problem. If I changed it to $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$ where $n$ takes on positive integers including $0$, would the answer change?

Answer 1

The function given by $f(x) := \frac{(-1)^x}{x}$ is defined only for $x \in \mathbb{Z}$ and $x\neq 0$, hence $\operatorname{Dom}(f) = \mathbb{Z} - \{ 0\}$.

The "points" at positive or negative infinity, i.e. $\pm \infty$, are both cluster points for $\operatorname{Dom}(f)$, therefore you can try to evaluate both $\displaystyle \lim_{x\to +\infty} f(x)$ and $\displaystyle \lim_{x\to -\infty} f(x)$.

Since $-\frac{1}{x} \leq f(x) \leq \frac{1}{x}$, both the aforementioned limits evaluate to $0$ by the Squeeze Theorem.

Answer 2

$(-1)^x$ for non-integer $x$ is "multi-valued" in the complex plane. But all values have the property $$ \big|(-1)^x\big| = |(-1)|^x = 1. $$ so, no matter what values you choose, for $x>0$ we have $$ \left|\frac{(-1)^x}{x}\right| = \frac{1}{x} $$

Answer 3

may you accept like below $$\forall x\to \infty :\lfloor x\rfloor \leq x \leq\lfloor x\rfloor+1$$now rewrite the limitation $$-1 \leq (-1)^{\lfloor x\rfloor} \leq 1\\ -1 \leq (-1)^{\lfloor x\rfloor+1} \leq 1\\\to \frac{-1}{x} \leq \frac{(-1)^x}{x}\leq \frac1x$$apply the limit to both sides

Answer 4

First, let's consider some terms of this sequence. Let $x\in\mathbb{N}\{1,2,3,...\}$, then $\frac{(-1)^x}{x}=(-1, \frac{1}{2},-\frac{1}{3},\frac{1}{4},...)$. Let's call the sequence $a_n=\frac{(-1)^x}{x}$.

Formally we define convergence of a sequence in this manner: "for all $\epsilon>0$, there exists some $n,M\in\mathbb{N}$ so that if $M>0$, then for all choices $n>M$ we have $|a_n-0|<\epsilon$. (NOTE: I have $0$ as the point inside the absolute value bars since this is the value we're trying to show is the limit of the sequence.)

I'm sorry if you're unfamiliar with $\epsilon$-proofs of a sequence. What this means is that if I choose any teeny-weeny number called $\epsilon$, I can find some other number $M$ that acts as a boundary for this sequence (think Gandalf from the "You shall not pass!" scene.) That way, if I have some other number $n$ so that $n>M$, then I'm guaranteed this sequence will be smaller than that teeny-weeny number $\epsilon$. (NOTE: this is why it's important to have a limit sign here. Obviously the sequence will never become $0$ because it is undefined at $0$, but it comes close enough so that we may say its limit is $0$.)

Now, for the fun part. Let's choose $\epsilon>0$ and $x\in\mathbb{N}$, then $|a_n-0|<\epsilon\Rightarrow|\frac{(-1)^x}{x}-0|<\epsilon\Rightarrow|\frac{1}{x}|<\epsilon\Rightarrow x>\frac{1}{\epsilon}$. We're going to choose $M=\frac{1}{\epsilon}$.

Now that we know our boundary point, we can say with confidence that if $x>M=\frac{1}{\epsilon}$, then it is always true that $|a_n-0|<\epsilon$, which is the same as saying $\lim_{x\rightarrow\infty}\frac{(-1)^x}{x}=0$.

Please don't hesitate to ask questions if this doesn't make sense.

Answer 5

To begin with when we say $\lim_{x\to \infty} a_x = K$, we are assuming that $x$ is restricted to a set only where $a_x$ is defined. As $(-1)^x$ is only defined on integers[*] to claim $\lim_{x\to \infty} \frac {(-1)^x}x=whatever$ we must assume $x \in \mathbb Z$. And as $x \to \infty$ we can limit ourselves to only considering $x \in \mathbb N$.

By convention "$x$" is a variable for a real number. If we are doing a limit restricted to $\mathbb N$ it is more conventional to use something like "$n$".

[**]I should also point out that even if we took a limit good on the reals, say $\lim_{x\to \infty} a_x = L$ then as $\mathbb N\subset \mathbb R$ then $\lim_{n\to \infty}a_n = L$ as well because if $x\to \infty; x\in \mathbb R \implies a_x \to L$ then $n\to \infty; n \in \mathbb N\subset \mathbb R$ so $n \in \mathbb R$ as well, the same result must follow.

Anyhow. If any $\epsilon > 0$ if $x > \frac 1{\epsilon}$ and $x\in \mathbb N$ then $(-1)^x = \pm 1$ and $|\frac {(-1)^x}x - 0| = |\pm \frac 1x| =|\frac 1x|=\frac 1x < \epsilon$. So that is the definition of $\lim_{x\to \infty} \frac {(-1)^x}x=0$.

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[*] Although $x^{3} = -1$ is solvable by $x=-1$ we do not define $(-1)^{\frac 13} = x$ because .... we just don't. there is no point in defining $b^x$ for $b < 0$ and all $x\in \mathbb R$ as such a function would not be consistent of $x\not \in \mathbb Z$, and wouldn't but continuous, and wouldn't work for any rational $\frac ab$ where $b$ is odd.

...unless you are using a text that does allow rationals with odd denominators. Some texts do. Oh well.

But in that case hee my paragraph marked [**].

Answer 6

In case you are in $\mathbb{N}$ ; $n \to \infty$ means:

either there is a $k$ such as $n = 2k$ with $k$ in $\mathbb{N}$ ($n$ is even.)

or there is a $k$ such as $n = 2k+1$ with $k$ in $\mathbb{N}$ ($n$ is odd.)

Moreover: when $n \to \infty$ you have certainly $n>1$

This means:

either $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{(-1)^{2k}}{k}$

or $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{(-1)^{2k+1}}{k+1}$

Which means

either $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{1}{k} = 0$

or $\lim_{n \to \infty} \frac{(-1)^n}{n}= lim_{k \to \infty} \frac{-1}{k+1} = 0$

Therefore $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$