Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $$f(x)=\frac{x}{x}$$ be defined on $\mathbb R\setminus \{0\}$. Show that $$\lim\limits_{x\to 0}f(x) = 1$$ without using l'Hospital's rule.

share|cite|improve this question
You do not need any approach... Just tell: how much is $x$ divided by $x$? – Godot Dec 6 '12 at 22:57

If $x \neq 0$, then $|f(x) - 1| = 0$. Let $\epsilon > 0$.
We need $\delta > 0$ so that $0<|x| < \delta\implies |f(x) - 1 |<\epsilon$ The value $\delta = 1$ works for any $\epsilon$.

share|cite|improve this answer
when you plug in x/x for f(x) you get 0 and in the definition it states that $ϵ > 0$ – Grigor Dec 6 '12 at 22:59
The definition of limit says $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$ Limits care about what happens around the point $a$ but are insensitive to what happens at $a$. – ncmathsadist Dec 6 '12 at 23:06

$$f(x)=1\qquad \forall x \neq 0$$

Thus $$f(1)=1$$ $$f(.001)=1$$ $$f(.00000000001)=1$$ etc.

You can get as close as you want to $x=0$ (without $x$ ever becoming $0$), and $f(x)$ will always be $1$.

share|cite|improve this answer
Does this answer truly deserve downvotes? If so, please elaborate. – Argon Dec 7 '12 at 0:27
I didn't downvote you, but you might want to add a bit more detail. It seems like the OP is confused about what limits mean, so you could try to explain that finding a limit is considering points very close to the limit. And so since one sees that $f(0.001)$ .... the limit is .... – Thomas Dec 7 '12 at 4:40

Is this a real question? $x/x = 1$ because $x \in {\mathbb R} \setminus \{0\}$, so ...

share|cite|improve this answer
yes, but how would you show that mathematically? – Grigor Dec 6 '12 at 22:56
@Grigor Mathematically, whenever $x\neq 0$; we have that $$\frac x x =1$$ Thus, on any punctured neighborhood of zero, $f(x)=1$ – Pedro Tamaroff Dec 6 '12 at 22:58
Gregor, it's definition, or better: it stems from definitions: for any real nonzero number $\,x\,$, it is true that $\,\frac{x}{x}=1\,$...we could get into equivalence classes and stuff, but I think the above should suffice. – DonAntonio Dec 6 '12 at 22:58
This is a tautology, $x/x = 1$ is always true. – glebovg Dec 6 '12 at 23:02
Perhaps at such a level as this question, one should assume that the author needs to see a delta-epsilon proof. – robjohn Dec 7 '12 at 2:15

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.