0

I am looking to prove that the function $f(x)=\frac{1}{x}$ is continuous at the point x=2. So we nee that given any $\epsilon>0,\ \exists\delta>0$ so that $|f(x)-f(2)|<\epsilon\\$ whenever $|x-2|<\delta$.

Scrtch Work:$|f(x)-f(2)|=|\frac{1}{x}-\frac{1}{2}|=|\frac{x-2}{2x}|=\frac{1}{2}\frac{|x-2|}{|x|}$. Now suppose, $|x-2|<\delta\leq\frac{1}{2}$, which implies $|x-2|<\frac{1}{2} \Rightarrow \frac{3}{2}<x<\frac{5}{2}$. Thus, $\frac{1}{2}\frac{|x-2|}{|x|}<(\frac{1}{2})(\frac{2}{3})\delta=\frac{\delta}{3}$. Here we want $\frac{\delta}{3}=\epsilon \Rightarrow \delta=3\epsilon$. And we're done if we take $\delta$=min{$\frac{1}{2},3\epsilon$}.

My question is, how do I work forward to show this argument works since I don't have a specific $\delta$?

Mr.Fry
  • 5,003
  • 3
  • 19
  • 28
  • You have a specific $\delta$, $\delta = \min {1/2, 3 \epsilon}$ –  Sep 13 '13 at 04:10
  • A related technique. – Mhenni Benghorbal Sep 13 '13 at 04:12
  • Yes. $\delta = \mathrm{min}(1/2, 3\varepsilon)$ is a specific number. All the $\varepsilon - \delta$ definition requires is that you show that a $\delta$ exists (and $0 < |x - c < \delta \implies |f(x) - L| < \varepsilon$). Your scratch work above shows that if $\delta = \mathrm{1/2, 3\varepsilon}$ then the definition is met, so a $\delta$ has been shown to always exist, as desired. –  Sep 13 '13 at 04:34

1 Answers1

0

Note that $$\left|\frac 1 x-\frac 1 a\right|=\frac{|x-a|}{|x||a|}$$

Now, we first want to make $x\neq 0$ in a nbhd of $a$. But $|x|\geqslant |a|-|x-a|>0$ if $|x-a|\leqslant \delta\leqslant |a|$. But then $$\frac{|x-a|}{|x||a|}\leqslant \frac{\delta}{|a|(|a|-\delta)}$$ for $|a|-|x|<|x-a|\leqslant \delta\implies |a|-\delta <|x|$.

With this, $$ \frac{\delta}{|a|(|a|-\delta)}\leqslant\varepsilon$$ gives $$\delta\leqslant\frac{|a|\varepsilon}{1+\varepsilon |a|}|a|<|a|$$ So $$\delta=\frac{|a|\varepsilon}{1+|a|\varepsilon}|a|$$ is a wise choice. I leave it to you to show this is actually the maximal solution, that is, any larger $\delta$ would fail to do the job. To do this, exhibit an $x'$ with $|x'-a|=\delta$ such that we get $=\varepsilon$.

Pedro
  • 122,002
  • @Crypto Maybe I completely misread your question. You asked about working "forward". I interpreted this as using our hypothesis to motivate a $\delta>0$ without an ansatz like $\leqslant \dfrac 1 2$. If this is not what you wanted, let me know. What you did is fine! ;) You even specified the $\delta$! Now I'm confused... – Pedro Sep 13 '13 at 04:30
  • @Crypto $$\left| {\frac{1}{x} - \frac{1}{2}} \right| = \left| {\frac{{x - 2}}{{2x}}} \right| = \frac{1}{2}\frac{{|x - 2|}}{{|x|}} < \frac{1}{2}\frac{2}{3}\delta = \frac{\delta }{3} = \frac{{3}}{3} = $$ is fine. Can you delete the code above? It is distorting the page. – Pedro Sep 13 '13 at 04:32
  • 1
    thanks a lot, this really helped – Mr.Fry Sep 13 '13 at 04:33