Michael Spivak, in his Calculus textbook pp 89, has tried to prove that given $\lim \limits_{x\to a} g(x)=m $, where $m\not=0$, $\lim _{x\rightarrow a} (\frac{1}{g})(x)=\frac{1}{m}$. To prove it, he states that if $\epsilon>0$ there is a $\delta$ such that for all $x$, if $0<|x-a|<\delta$, then $|g(x)-m|< \min\left(\frac{|m|}{2},\frac{\epsilon|m|^2}{2}\right)$.I do not understand how he gets/states that.Can you please help me prove it?He later uses a lemmma to prove the theorem but I understand the proof but not the intermediate statement. Thanks in advance.


The fact that $$\lim_{x\to a}g(x) = m$$ means: for every $\varepsilon\gt 0$, there exists $\delta\gt 0$ such that if $0\lt |x-a|\lt \delta$, then $|g(x) -m|\lt \varepsilon$.

You can pick whatever $\varepsilon\gt 0$ you want, there will be a corresponding $\delta$.

Given $\epsilon\gt 0$, the number $$\min\left(\frac{|m|}{2},\frac{\epsilon|m|^2}{2}\right)$$ is a positive number; so letting it be $\varepsilon$, we know there exists $\delta\gt 0$ with the desired property.

So that's how he "gets/states that". It's a consequence of the limit $$\lim_{x\to a}g(x) = m.$$ Of course, you might wonder why he chooses that particular $\varepsilon$. The reason is that it makes the proof work. Intuitively: you want to make sure that $g(x)$ is close enough to $m$ so that $\frac{1}{g(x)}$ is close enough to $\frac{1}{m}$. If you work out wha tyou need to ensure that, the condition will become apparent.

  • $\begingroup$ Thank you.I think I now understand it . $\endgroup$ – Eisen Nov 10 '11 at 6:35
  • $\begingroup$ Arturo, Not sure your post answers the question. The key reason why Spivak's method is a proof is lacking, which is that $K(m,\epsilon)=\min(\frac12|m|,\frac12\epsilon|m|^2)$ can be made as small as desired, for suitable values of $\epsilon$ (in short $K(m,\epsilon)\to0$ when $\epsilon\to0$). $\endgroup$ – Did Nov 10 '11 at 7:29
  • $\begingroup$ @DidierPiau: I'm not sure I agree; that would make sense if Spivak were trying to show that the limit of $g(x)$ is $m$; but he isn't. He is showing that the limit of $\frac{1}{g}$ is $\frac{1}{m}$. In the proof, $\epsilon$ is given, and the fact that $\lim g(x)=m$ is given; he makes the choices of $\delta$ to ensure that you get $|\frac{1}{g}(x)-\frac{1}{m}|\lt\epsilon$, and one of the auxiliary choices he makes uses $K(m,\epsilon)$. It really doesn't matter that $K(m,\epsilon)\to 0$ for suitable choices, since he is not trying to show $|g(x)-m|$ can be made arbitrarily small. $\endgroup$ – Arturo Magidin Nov 10 '11 at 14:22
  • $\begingroup$ In other words, the idea of the proof is to show that (1) for every $\epsilon$, there exists $\delta$ such that $|x-a|<\delta$ implies $|g(x)-m|<K(m,\epsilon)$ and (2) for every $y$ ands $\epsilon$, $|y-m|<K(m,\epsilon)$ implies that $|1/y-1/m|<\epsilon$. Hmmm... Then my previous comment was off topic. Sorry about that and thanks for the explanation. $\endgroup$ – Did Nov 10 '11 at 16:34
  • $\begingroup$ @DidierPiau: No problem! You're right that one has to be careful; there are two ways to go about these limits: show that for every $\epsilon\gt 0$ you can find $K(\epsilon)$ and $\delta\gt 0$ such that $|x-a|\lt\delta$ implies $|f(x)-m|\lt K(\epsilon)$ and $K(\epsilon)\to 0$ for suitable choices of $\epsilon$; or make inspired choices of $\delta$ that ensure some auxiliary term is less than $K(\epsilon)$ and that this implies $|f(x)-m|\lt \epsilon$. Spivak generally sticks to the latter rather than the former, but when we have the former one should definitely emphasize the $K(\epsilon)\to 0$. $\endgroup$ – Arturo Magidin Nov 10 '11 at 16:57

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