# Thread: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exist

1. ## Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exist

Show that $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.

Is my following argument correct?

I will show there exists a sequence $(x_n) \subset \mathbb{R} \backslash \{0\}$ satisfying $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $\lim_{n \rightarrow \infty} f(x_n)$ does not exist. Consider $(x_n) = \frac{1}{n}$ for $n \in \mathbb{N}$, clearly $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $f(x_n) = n$, which diverges to $+\infty$ as $n \rightarrow \infty$, hence $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.

2. ## Re: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exi

Originally Posted by usagi_killer
Show that $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.

Is my following argument correct?

I will show there exists a sequence $(x_n) \subset \mathbb{R} \backslash \{0\}$ satisfying $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $\lim_{n \rightarrow \infty} f(x_n)$ does not exist. Consider $(x_n) = \frac{1}{n}$ for $n \in \mathbb{N}$, clearly $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $f(x_n) = n$, which diverges to $+\infty$ as $n \rightarrow \infty$, hence $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.
I've never heard the term "Sequential definition of a limit" ,but I do know that in order to show a limit does not exist you must show that the limit approaching the value from the right is different from the limit approaching the value from the left i.e $\displaystyle \lim_{x\rightarrow a^{-}} f(x) \neq \lim_{x\rightarrow a^{+}} f(x)$ I Could be wrong about this, but this has been my experience.

On the first line you say that "$\displaystyle \lim_{n\rightarrow\infty}f(x_n)\quad\text{DNE}$ ";however, in the next line you say " $\displaystyle \lim_{n\rightarrow\infty} f(x_n)\rightarrow\infty$ " This is a contradiction. I have the feeling in the first line you meant to say "but $\displaystyle \lim_{n\rightarrow\infty}f(x_n)\rightarrow\infty$"

I also have to raise the question: where did you get $\displaystyle f(x_n)=n$ from?

3. ## Re: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exi

Hi,
I agree that your argument shows that there is no finite limit. However, you still need to show that neither $+\infty$ nor $-\infty$ is the limit. You can modify your argument to handle these two cases.

4. ## Re: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exi

What he calls the sequential definition of a limit is what I call the sequential criterion for convergence of a function: the theorem $\displaystyle \lim_{x\to{a}} f(x) = L$ if and only if for all sequences $\displaystyle (x_n)$ converging to a, the sequence $\displaystyle (f(x_n))$ converges to L. Some authors use this as the definition of the limit of a function and prove the epsilon-delta definition as a theorem.

Anyway, what johng says is exactly correct. Since your $\displaystyle x_n$ is always positive, if you look at the graph of $\displaystyle \frac{1}{x}$, you're only looking at the part of the function in the first quadrant. That's why your argument doesn't exclude $\displaystyle +\infty$ as a possible limit - if you only look in the first quadrant, that's what it looks like the function does. So choose a sequence that has positive and negative values.

- Hollywood

5. ## Re: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exi

How about this argument then? Define $f: \mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}, f(x) = 1/x$. I claim that $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist. I will use the divergence criterion for functional limits to show this. Define $(x_n) = 1/n$ and $(y_n) = -1/n$, $n \in \mathbb{N}$ which are both sequences in $\mathbb{R} \backslash \{0\}$. Note that $x_n \neq 0$ and $y_n \neq 0$ for all $n$. Furthermore, $f(x_n) = n$ and $f(y_n) = - n$. Observe that $\lim x_n = \lim y_n = 0$, but $\lim f(x_n) = + \infty$ and $\lim f(y_n) = - \infty$, hence $\lim f(x_n) \neq \lim f(y_n)$, thus $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.

6. ## Re: Sequential definition of a limit to show limit as x goes to 0 of 1/x does not exi

Hi again,
I completely accept this argument.