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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0when you plug in x/x for f(x) you get 0 and in the definition it states that $ϵ > 0$ – 2012-12-06
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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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2yes, but how would you show that mathematically? – 2012-12-06
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2@Grigor *Mathematically*, whenever $x\neq 0$; we have that $$\frac x x =1$$ Thus, on any punctured neighborhood of zero, $f(x)=1$ – 2012-12-06
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0Gregor, 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. – 2012-12-06
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0This is a tautology, $x/x = 1$ is always true. – 2012-12-06
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0Who downvoted my answer? What is the point of downvoting the right answer? – 2012-12-07
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0@glebovg: $x/x=1$ except when $x=0$ – 2012-12-07
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0@robjohn I just deleted my reply to Snowball because he deleted his comment, but the question states that $x \in {\mathbb R} \setminus \{0\}$ and I did not want to type it again. Stop downvoting my answer and read the question first! – 2012-12-07
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0@glebovg: I have not downvoted a single post, so I certainly did not downvote yours. Since the question is about the limit as $x\to0$, I like to make sure that $x=0$ is excluded. I was just making sure that no one would interpret your statement at $x=0$. – 2012-12-07
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0OK. I just do not understand why people downvote a correct answer. – 2012-12-07
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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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0Does this answer truly deserve downvotes? If so, please elaborate. – 2012-12-07
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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