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

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In summary, the conversation discusses how to show that the limit of $\frac{1}{x}$ as $x$ approaches $0$ does not exist. One way to do this is by showing that 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. Another approach is to consider the behavior of $\frac{1}{x}$ near $0$ and observe that it is positively or negatively unlimited, leading to the conclusion that the limit does not exist. However, authors may differ on
  • #1
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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  • #2
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.
 
  • #3
Yup that is one way, however I was wondering whether my proof using the sequential characterisation is correct?
 
  • #4
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.
 
  • #5
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

$\chi$ $\sigma$
 
  • #6
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.
 

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

What does it mean for the limit of x to 0 of 1/x to not exist?

When we say that the limit of x to 0 of 1/x does not exist, it means that there is no single value that the function approaches as x gets closer and closer to 0. This could happen because the function is undefined or oscillates between different values near 0.

How do we prove that the limit of x to 0 of 1/x does not exist?

In order to prove that the limit of x to 0 of 1/x does not exist, we need to show that for any value L, there exists an interval around 0 where the function values are either undefined or do not approach L as x gets closer to 0. This can be done through calculus techniques such as the squeeze theorem or using the definition of a limit.

Can the limit of x to 0 of 1/x approach infinity?

No, the limit of x to 0 of 1/x does not approach infinity. This is because as x gets closer to 0, the function values get larger and larger, but they never reach a specific value or approach infinity. Therefore, the limit does not exist.

Is it possible for the limit of x to 0 of 1/x to be negative?

No, the limit of x to 0 of 1/x cannot be negative. This is because the function values for 1/x are always positive, no matter how close x gets to 0. Therefore, the limit can only approach positive infinity or not exist at all.

Can we use a graph to show that the limit of x to 0 of 1/x does not exist?

Yes, we can use a graph to visually show that the limit of x to 0 of 1/x does not exist. The graph will have a vertical asymptote at x=0, indicating that the function values become infinitely large as x approaches 0. This confirms that the limit does not exist.

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