# Show that Let $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ be defined by $f(x) = \frac{1}{x}$. Show $\lim_{x \to 0}\frac{1}{x}$ doesn't exist.

Show that Let $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ be defined by $f(x) = \frac{1}{x}$. Show $$\lim_{x \to 0}f(x)$$ doesn't exist.

I don't think I've ever seen a rigorous proof of this. I'm trying to show rigorously that this limit doesn't exist.

My Attempted Proof: Observe that $0$ is a limit point of $\mathbb{R} \setminus \{0\}$ so it makes sense to talk about the limit of $f(x)$ as $x \to 0$. Suppose that $$\lim_{x \to 0} f(x) = q \in \mathbb{R}.$$ Then for all $\epsilon > 0$ there exists a $\delta > 0$ such that $|x- 0|< \delta \implies \left|\frac{1}{x} - q\right| < \epsilon$.

Now this is where I got stuck, I want to show that a contradiction arises from the above which will allow me to conclude that no such $q \in \mathbb{R}$ exists. The only way to do this (I think) would be to find an $\epsilon > 0$ such that for all $\delta > 0$ we have $|x| < \delta$ and $\left|\frac{1}{x} - q\right| > \epsilon$. I'm having trouble finding such an $\epsilon$. The only candidate I can think of is $\epsilon = \frac{1}{|q|}$ but I'm still not sure how two conclude that $\left|\frac{1}{x} - q\right| > \frac{1}{|q|}$.

• Let $\epsilon$ be 1 and then no matter what delta is, you can pick $x$ to be the min($\frac{1}{q+2},\delta$). Thus $|x| \leq \delta$ and $|\frac{1}{x}-q| \geq \epsilon$. (This is assuming q to be positive, but something very similar works in the case of q negative.) – JonHales Jun 20 '18 at 13:59

Suppose that $f: U → R$ is an application defined on a subset $U$ of the set $R$ of reals. If $p$ is a real, not necessarily belonging to $U$ but such that $f$ is "defined in the neighborhood of $p$", we say that $f$ admits a limit (finite) at point $p$, if there exists a real $L$ satisfying:

for every real $ε> 0$ there exists a real $δ> 0$ such that for all $x$ in $U$ such that $| x - p | <δ$, we have $| f (x) - L | <ε$.

We can take $ε = 1$, there exists a real $δ'> 0$ such that for all $x$ in $U$ such that $| x - p | <δ$, we have $| f (x) - L | <1$, which is equivalent to $L-1<f(x) < L+1$.

So if $f$ has a finite limit in $0$, there must exists a neighborhood of $0$ such that the value taken by $f$ in this neighborhood is bounded.

In the case here, the serie $(f(1/n))_{n\in\mathbb{N}}$ converges to $+\infty$, so f doesn't have a finite limit in $0$.

• Could you explain your answer in a bit more detail? For example is there a theorem that states that if $f$ has a finite limit in $0$, then the value taken by $f$ in the neighborhood of $0$ is bounded? – Perturbative Jun 20 '18 at 14:00

You are right… take any $\varepsilon < 1$ and any $\delta > 0$

Then choose $n \ge q+1$ such that $x = \frac{1}{n} < \delta$ and then you have $$|x| < \delta$$ as well as $$\left|\frac{1}{x} - q\right| \ge 1 > \varepsilon$$ and you are done.

You can prove that the limit is not $q$ by proving that, for each $\delta>0$, there is a $x\in(-\delta,\delta)\setminus\{0\}$ such that $\left|\frac1x-q\right|\geqslant1$. In order to do that, assuming that $q\geqslant 0$, take $x\in(0,\delta)$ such that $\frac1x\geqslant1+q(\iff x\leqslant\frac1{1+q}$). The proof is similar if $q\leqslant0$.

In this case, you don't have to search for an $\varepsilon$, because $$\lim_{x\to 0^+}\frac{1}{x}=+\infty$$ and $$\lim_{x\to 0^-}\frac{1}{x}=-\infty.$$

You can verify that for all $M>0$ you can find a $\delta>0$ such that for every $x|0<x<\delta$ then $\frac{1}{x}>M$, and for every $x|-\delta<x<0$ then $\frac{1}{x}<-M$.

For the unicity of the limit,then $\lim_{x\to0}\frac{1}{x}$ doesn't exist.

1. Non finite limit exists. Otherwise, since $xf(x)=1$ for every $x \neq 0$, you would get $1=\lim_{x \to 0} 1 = \lim_{x \to 0} x f(x) = 0 \cdot \lim_{x \to 0} f(x)=0$, a contradiction.
2. No infinite limit exists. Suppose that $\lim_{x \to 0} f(x)=+\infty$. In particular, in a suitably small (punctured) neighborhood of $0$, the function $f$ would be always strictly positive. This is false, since $1/x<0$ for $x<0$. In the same way, you can exclude that $\lim_{x \to 0} f(x)=-\infty$.

Anyway, this is almost a divertissement. You should learn soon that $$\lim_{x \to 0-} \frac{1}{x}=-\infty, \qquad \lim_{x \to 0+} \frac{1}{x}=+\infty,$$ so that the limit as $x \to 0$ does not exist.

Consider

1) $(x_n)=1/n , n \in.\mathbb{Z^+}$.

$\lim_{n \rightarrow \infty}x_n= 0.$

$\lim_{ n \rightarrow \infty} f(x_n) =$

$\lim_{n \rightarrow \infty} n =\infty .$

2) Consider $(x_n) =-1/n$, $n\in \mathbb{Z^+}$.

$\lim_{n \rightarrow \infty } x_n =0.$

$\lim_{n \rightarrow \infty }f(x_n)=$

$\lim_{n \rightarrow \infty} (-n )=-\infty.$

Does the limit exist?

P.S. Can you show that $\lim_{n \rightarrow \infty} x_n=0$ ?