# Show that $f(x)=1/ x$ is continuous at any $c\neq 0$

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

• Can 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 8 '13 at 6:51
• do you mean the definition of continuous? thanks – leena adam Apr 8 '13 at 6:58
• For any $x$ such that $|x - c| < \delta$, $|\tfrac{1}{x} - \tfrac{1}{c}| < \varepsilon$. – Sammy Black Apr 8 '13 at 6:59
• This question seems like a duplicate of math.stackexchange.com/q/12462/264 – Zev Chonoles Apr 8 '13 at 7:01
• Play with the $\varepsilon$ inequality to try to determine how to choose $\delta$ (in terms of $\varepsilon$) to force the inequality. – Sammy Black Apr 8 '13 at 7:02

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.
• thank 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 8 '13 at 7:24
• Try choosing $\delta < \min(\frac{c^2\epsilon }{2},\frac{|c|}{2})$ and see how that works with the above. – copper.hat Apr 8 '13 at 7:48
• @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