-3
$\begingroup$

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.

  • 3
    You do not need any approach... Just tell: how much is $x$ divided by $x$?2012-12-06

3 Answers 3

5

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 $\epsilon$.

  • 3
    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 $a$.2012-12-06
2

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

  • 3
    Perhaps at such a level as this question, one should assume that the author needs to see a delta-epsilon proof.2012-12-07
1

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

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 $1$.

  • 0
    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 ....2012-12-07