# Is $\lim _{x\rightarrow 0}\dfrac {1} {x}=\infty$ right?

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!

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Your understanding of the matter is correct: the two one-sided limits are different. –  Brian M. Scott Feb 15 '13 at 11:33
You should tag it as homework I guess. –  Marra Feb 15 '13 at 11:33
@Gustavo Marra Thank you for your suggestion. –  felix34 Feb 15 '13 at 11:36

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 $-\infty$.

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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.

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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 - http://en.wikipedia.org/wiki/Limit_of_a_function#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.

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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 $-\infty$.

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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.

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