Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just learned a little about the limit by myself, and I wonder the result of $\lim _{x\rightarrow 0}\dfrac {1} {x}$.

In order to get the answer, I asked one of my friends, and he told me that it is equal to $\infty$:

$$\lim _{x\rightarrow 0}\dfrac {1} {x}=\infty$$

But I was still puzzled.

In my opinion, the variable $x$ can approach $0$ from both positive direction and negative direction. So I get $\lim _{x\rightarrow 0^+}\dfrac {1} {x}=+\infty$ and $\lim _{x\rightarrow 0^-}\dfrac {1} {x}=-\infty$.

Could you tell me your ideas about the result of $\lim _{x\rightarrow 0}\dfrac {1} {x}$?

Thanks a lot!

share|cite|improve this question
Your understanding of the matter is correct: the two one-sided limits are different. – Brian M. Scott Feb 15 '13 at 11:33
up vote 4 down vote accepted

Certainly $\displaystyle\lim_{x\to0}\frac1x=\infty$ within the projective line $\mathbb R\cup\{\infty\}$. But when one works in another conventional space $\mathbb R\cup\{\pm\infty\}$ then one of the one-sided limits is $+\infty$ and the other is $-\infty$.

share|cite|improve this answer

If the two one-sided limits of a function at a point are different, that is, if

$$\lim_{x\to p^+} f(x) \ne \lim_{x\to p^-} f(x)$$

then the limit of the function $f$ isn't defined at point $p$, precisely because it can take on two different values depending on how one goes about it.

This means that, whilst it is true that

$$\lim_{x\to 0^+} \frac{1}{x} = \infty, \quad \lim_{x\to 0^-} \frac{1}{x} = -\infty$$

The limit of $\frac{1}{x}$ at $x = 0$ doesn't exist.

share|cite|improve this answer
if the one-sided limits are different then the function can very well be defined at $p$, it's just that the limit at $p$ does not exist. The particular value of a function at $p$, or indeed whether or not the function is defined at $p$ is never of any consequence to questions concerning the limits of the function at $p$. – Ittay Weiss Feb 15 '13 at 11:46
Of course, and the answer is edited accordingly. I could've sworn I had the words 'limit of' before that statement, but clearly did not. Thanks for noticing! – tesc Feb 15 '13 at 11:50
my pleasure :) :) – Ittay Weiss Feb 15 '13 at 11:52

$\lim_{x\to x_{0}}f\left(x\right)=\infty\Longleftrightarrow\lim_{x\to x_{0}}\left|f\left(x\right)\right|=+\infty $
Unsigned infinity as infinite distance from $0$ (or any real number) - in both directions. Signed infinity as infinite distance in only direction.

Wiki calls this alternative notation -

I learned about it by such definition:

neighborhood of $U_\epsilon(\infty)$ is a set of the form ${x: |x|>\dfrac{1}{\epsilon}}$

Using this definition, $f(x)$ gets more and more close to $\infty$ (when $x\to 0$).
And the limit $\lim _{x\rightarrow 0}\dfrac {1} {x}=\infty$ is calculated correctly.

share|cite|improve this answer

The definition of limits is usually such that a limit only exists if it is the same from all directions of approach to a point. Thus, since you get a different limit from each side, the limit does not exist. This is, of course, assuming you are looking for the limit over $\mathbb{R}$, as the limit over $\mathbb{R}^{+}$ or $\mathbb{R}^{-}$ can be defined as a one-sided limit, and thus can exist as you have stated above, as either $\infty$ or $-\infty$.

share|cite|improve this answer
Welcome to MSE! I have reformatted slightly your answer, to adhere to the LaTeX conventions used here. Please take note for your further contributions. – Andreas Caranti Feb 15 '13 at 11:39

If $x\rightarrow 0^-$ then the limit is $-\infty$. If $x\rightarrow 0^+$ then the limit is $+\infty$. It's not hard to show this through the definition of limith though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.