Show that limit of x to 0 of 1/x does not exist

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Discussion Overview

The discussion revolves around the limit of the function \( \frac{1}{x} \) as \( x \) approaches 0, specifically addressing whether this limit exists. Participants explore various arguments and interpretations related to the limit's behavior from both the left and right sides, as well as the implications of using infinity in limit statements.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants present a sequence \( (x_n) = \frac{1}{n} \) to demonstrate that as \( n \) approaches infinity, \( f(x_n) \) diverges to \( +\infty \), suggesting that the limit does not exist.
  • Another participant notes that the left-hand limit \( \lim_{x \to 0^-} \frac{1}{x} = -\infty \) and the right-hand limit \( \lim_{x \to 0^+} \frac{1}{x} = +\infty \) are not equal, reinforcing the argument that the limit does not exist.
  • One participant cautions against the interpretation of limits approaching \( +\infty \) or \( -\infty \) as definitive values, arguing that these are not numbers and thus complicate the discussion about the existence of limits.
  • Another participant suggests that while some authors may assert that the limit does not exist, others may express it as tending to infinity, emphasizing the need for clarity in definitions and interpretations.
  • A later reply introduces the limit of \( \frac{1}{x^2} \) as an example where some authors may claim it does not exist, while others might assert it approaches infinity, highlighting the variability in interpretations of limits involving infinity.

Areas of Agreement / Disagreement

Participants express differing views on the existence of the limit of \( \frac{1}{x} \) as \( x \) approaches 0. While some agree that the limit does not exist based on the behavior of the function, others raise concerns about the implications of using infinity in limit statements, indicating that the discussion remains unresolved.

Contextual Notes

Participants highlight the importance of careful language when discussing limits involving infinity, noting that such terms can lead to misunderstandings. The discussion reflects a range of interpretations and definitions regarding limits and their existence, particularly in the context of extended real number systems.

Usagi
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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.
 
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Usagi said:
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.

$\displaystyle \begin{align*} \lim_{x \to 0^-} \frac{1}{x} = -\infty \end{align*}$, but $\displaystyle \begin{align*} \lim_{x \to 0^+} \frac{1}{x} = +\infty \end{align*}$. Since these aren't the same, the limit does not exist.
 
Yup that is one way, however I was wondering whether my proof using the sequential characterisation is correct?
 
Authors can differ on this matter. It's one thing to say that:

$\displaystyle \lim_{x \to 0} f(x) = \infty$

and quite another to say that the limit does not exist.
 
In my [very modest] care must be used in pronuncing sentences like $\displaystyle \lim_{x \rightarrow 0} f(x) = + \infty$ or $\displaystyle \lim_{x \rightarrow 0} f(x) = - \infty$ and the reason is easily to understand. Neither $+ \infty$ nor $- \infty$ are numbers so that such sentences, if literally understood, are nonsenses. The reason why these sentences are commonly used in practice is that they are a short form to say that f(x) is positively or negatively unlimited near x=0. The reason why $\displaystyle \lim_{x \rightarrow 0} \frac{1}{x}$ doesn't exist has been correctly explained by Usagi in the first post. If absurdly a real number l for which $\displaystyle \lim_{x \rightarrow 0} \frac{1}{x}= l$ exists, then fon any $\varepsilon> 0$ a $\delta>0$ exists for which is $|l - \frac{1}{\delta}|< \varepsilon$. But $\|\frac{1}{x}|$ is unlimited near x=0, so that all that cannot be true...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
In my [very modest] care must be used in pronuncing sentences like $\displaystyle \lim_{x \rightarrow 0} f(x) = + \infty$ or $\displaystyle \lim_{x \rightarrow 0} f(x) = - \infty$ and the reason is easily to understand. Neither $+ \infty$ nor $- \infty$ are numbers so that such sentences, if literally understood, are nonsenses. The reason why these sentences are commonly used in practice is that they are a short form to say that f(x) is positively or negatively unlimited near x=0. The reason why $\displaystyle \lim_{x \rightarrow 0} \frac{1}{x}$ doesn't exist has been correctly explained by Usagi in the first post. If absurdly a real number l for which $\displaystyle \lim_{x \rightarrow 0} \frac{1}{x}= l$ exists, then fon any $\varepsilon> 0$ a $\delta>0$ exists for which is $|l - \frac{1}{\delta}|< \varepsilon$. But $\|\frac{1}{x}|$ is unlimited near x=0, so that all that cannot be true...

Kind regards

$\chi$ $\sigma$

Right, but we could remedy this situation by considering:

$\displaystyle \lim_{x \to 0} \dfrac{1}{x^2}$.

Some authors will say this limit does not exist (it IS true that there is no real number it tends to), where others will write:

$\displaystyle \lim_{x \to 0} \dfrac{1}{x^2} = \infty$

where the forgoing is understood to mean:

$\forall N \in \Bbb N, \exists \delta > 0: 0 < |x| < \delta \implies \dfrac{1}{x^2} > N$,

a perfectly sound statement that does not even mention "infinity" (or what such a term may mean).

Alternatively, one can work within "an extended real number system" $\Bbb R \cup \{-\infty,\infty\}$, where one has to modify the field axioms for the non-finite special cases (in general, we have to avoid various "indeterminate forms" such as:

$\dfrac{\infty}{\infty}$ or $\infty - \infty$).

Lots of strange situations can arise by "treating infinity as a number", so it is wise to be very specific about what one intends to convey.
 

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