Can anyone help me to solve this problem?
Q: Show that $f:\mathbb{R}\rightarrow\mathbb{R}$, $f(x)=1/ x$ is continuous at any $c\neq 0$.
Notice: (choose your $\delta$ so that you stay away from 0)
I hope someone can solve.
Thanks
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scjorge
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leena adam
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2Can you include what you've tried? For example, write down what continuity of $f$ at $c$ means in terms of $\delta$ and $\varepsilon$. – Sammy Black Apr 08 '13 at 06:51
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do you mean the definition of continuous? thanks – leena adam Apr 08 '13 at 06:58
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For any $x$ such that $|x - c| < \delta$, $|\tfrac{1}{x} - \tfrac{1}{c}| < \varepsilon$. – Sammy Black Apr 08 '13 at 06:59
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@leenaadam: Yes, that is just a different way of saying the same thing. Do you know the definition of continuous? Try writing down what that definition says, replacing in the definition your particular function $f(x)=\frac{1}{x}$. – Zev Chonoles Apr 08 '13 at 07:00
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1This question seems like a duplicate of http://math.stackexchange.com/q/12462/264 – Zev Chonoles Apr 08 '13 at 07:01
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1Play with the $\varepsilon$ inequality to try to determine how to choose $\delta$ (in terms of $\varepsilon$) to force the inequality. – Sammy Black Apr 08 '13 at 07:02
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I recently posted an answer at math.stackexchange.com/q/12462/264, so you should just take a look at that answer. – Angel Oct 22 '21 at 15:49
2 Answers
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Suppose $c>0$. If $x > \frac{c}{2}$, then $|\frac{1}{x}-\frac{1}{c}|= \frac{|x-c|}{xc} < \frac{2}{c^2}|x-c|$
So, if $\epsilon>0$, choose $\delta < \min(\frac{c^2\epsilon }{2},\frac{c}{2})$, then if $|x-c|< \delta$, we have (i) $x > \frac{c}{2}$, and from above we have (ii) $|\frac{1}{x}-\frac{1}{c}| < \frac{2}{c^2}\delta \le \epsilon $.
A similar argument applies to $c<0$. Or you could use the fact that $f(x) = -f(-x)$, and use the fact that multiplication by a constant ($-1$, in this case) is continuous, and composition of continuous functions is continuous.
copper.hat
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1thank you very much Mr copper can you please apply when c<0 because this part little hard to me and how much become the value of $\delta$ thank you so much – leena adam Apr 08 '13 at 07:24
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2Try choosing $\delta < \min(\frac{c^2\epsilon }{2},\frac{|c|}{2})$ and see how that works with the above. – copper.hat Apr 08 '13 at 07:48
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No, not the answer but the question makes me wonder so, actually i am confused if by definition 1/|x| is continuous or not – Display name Mar 17 '18 at 08:11
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3@Displayname: The question is ill posed, as such. The function $f$ is not defined at $x=0$. – copper.hat Mar 17 '18 at 17:46
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@joseph: It doesn't have to be true, but if it is true, then I can show that $|{1 \over x} - {1 \over c}|$ is bounded by some multiple of $|x-c|$. – copper.hat Oct 25 '19 at 23:13
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1@joseph: I don't understand why you are asking. If it is not true then the subsequent condition is not guaranteed to hold. – copper.hat Oct 25 '19 at 23:18
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@AntonioMariaDiMauro: I don't understand your question, could you elaborate please? – copper.hat Mar 04 '20 at 22:41
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This is can be solved without considering two cases $c<0$ and $c>0$. Because, we have $|\frac{1}{x}-\frac{1}{c}|=\frac{|x-c|}{|x||c|}$. We need find a positive lower bound for $|x|$, it can be done as $x$ is non-zero so $|x|$ should lie right to $0$.
Choosing $|x-c|<|c|/2$, we have $|c|=|c-x+x|\leq |x-c|+|x|\leq |c|/2+|x|$.
$|c|-|c|/2 \leq |x| \implies |c|/2\leq |x|\implies 1/|x|\leq 2/|c|$. This completes the proof.
Madhan Kumar
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