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.
Proof of Spivak's statement that if 0<|x-a|<\delta, then |g(x)-m|< \min(\frac{|m|}{2},\frac{\epsilon|m|^2}{2})
-
2You can see the link http://math.stackexchange.com/questions/54754/proving-that-b-n-to-b-implies-left-frac1b-n-right-to-frac1b that may help you. – 2011-11-10
1 Answers
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.
-
0@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 $K(\epsilon)\to 0$. – 2011-11-10