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.
Evaluate the limit without l'Hospital's rule
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calculus
limits
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3You do not need any approach... Just tell: how much is $x$ divided by $x$? – 2012-12-06
3 Answers
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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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3The 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
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Is this a real question? $x/x = 1$ because $x \in {\mathbb R} \setminus \{0\}$, so ...
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3Perhaps at such a level as this question, one should assume that the author needs to see a delta-epsilon proof. – 2012-12-07
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$f(x)=1\qquad \forall x \neq 0$
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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0I 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