# Evaluate the limit without l'Hospital's rule

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.

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

## 3 Answers

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 $\epsilon$.

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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

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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 $1$.

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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 ...

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yes, but how would you show that mathematically? – Grigor Dec 6 '12 at 22:56

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@Grigor Mathematically, whenever $x\neq 0$; we have that $$\frac x x =1$$ Thus, on any punctured neighborhood of zero, $f(x)=1$ – Peter 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

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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
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