Lemma that is used for L'hospital's rule

• Apr 10th 2011, 12:08 PM
paulrb
Lemma that is used for L'hospital's rule
This proof is left as an exercise in the book:

Suppose that g is continuously differentiable in an interval about $\displaystyle x_0$ and $\displaystyle g(x_0) = 0$. If $\displaystyle g'(x_0) \neq 0$, show that there is a small interval about $\displaystyle x_0$ in which neither g nor g' vanish except for the root of g at the point $\displaystyle x_0$.

I don't really know what this is saying, nor how to prove to prove it.
• Apr 10th 2011, 12:13 PM
Drexel28
Quote:

Originally Posted by paulrb
This proof is left as an exercise in the book:

Suppose that g is continuously differentiable in an interval about $\displaystyle x_0$ and $\displaystyle g(x_0) = 0$. If $\displaystyle g'(x_0) \neq 0$, show that there is a small interval about $\displaystyle x_0$ in which neither g nor g' vanish except for the root of g at the point $\displaystyle x_0$.

I don't really know what this is saying, nor how to prove to prove it.

What don't you understand about the statement? It's saying suppose that you found some point $\displaystyle x_0$ such that $\displaystyle g(x_0)=0$ yet $\displaystyle g'(x_0)\ne 0$ can you find some neighborhood of $\displaystyle x_0$ such that $\displaystyle g'$ and $\displaystyle g$ are neither zero anywhere on that neighborhood except $\displaystyle g$ is of course zero at $\displaystyle x_0$. Anyways..

Since $\displaystyle g'(x_0)\ne 0$ and $\displaystyle g'$ is continuous there exists an open ball $\displaystyle B_\delta(x_0)$ such that $\displaystyle g'$ does not vanish on that open ball. Now, if $\displaystyle g$ vanished somewhere else on that open ball besides $\displaystyle x_0$ you could apply Rolle's theorem to find a zero of $\displaystyle g'$ on the open ball contradictory to construction.
• Apr 10th 2011, 12:30 PM
paulrb
Thanks...although we have not covered what an open ball is in the class. We have covered Rolle's theorem though.

Intuitively I think I understand now. Since $\displaystyle g'(x_0) \neq 0$, there has to be some point near $\displaystyle x_0$where g is not 0. Since g is continuous, all of the points between $\displaystyle x_0$ and this point are nonzero. Furthermore for g to reach this nonzero point from 0, there has to be an interval between them where g'(x) is nonzero, since g'(x) is continuous. Although I still don't know how to prove this rigorously.
• Apr 10th 2011, 12:37 PM
Drexel28
Quote:

Originally Posted by paulrb
Thanks...although we have not covered what an open ball is in the class. We have covered Rolle's theorem though.

Intuitively I think I understand now. Since $\displaystyle g'(x_0) \neq 0$, there has to be some point near $\displaystyle x_0$where g is not 0. Since g is continuous, all of the points between $\displaystyle x_0$ and this point are nonzero. Furthermore for g to reach this nonzero point from 0, there has to be an interval between them where g'(x) is nonzero, since g'(x) is continuous. Although I still don't know how to prove this rigorously.

Hmm, I think you're mixing up the argument a little bit. Let me rephrase my argument in nicer language. We know that $\displaystyle g'(x_0)\ne 0$, right? Let's assume just for ease that $\displaystyle g'(x_0)>0$. Then, by continuity of $\displaystyle g'$ there exists some $\displaystyle \delta>0$ such that $\displaystyle \displaystyle |x-x_0|<\delta\implies \left|g(x_0)-g(x)\right|<\frac{g(x_0)}{2}$ which in particular implies that if $\displaystyle |x-x_0|<\delta$ then $\displaystyle g'(x)\ne 0$ (check this). Thus, we have on the interval $\displaystyle (x_0-\delta,x_0+\delta)$ that $\displaystyle g'$ is not zero. Now, I claim that except for $\displaystyle x_0$ $\displaystyle g(x)\ne 0$ for $\displaystyle x\in(x_0-\delta,x_0+\delta)$. Why? If $\displaystyle x<x_0\in (x_0-\delta,x_0+\delta)$ such that $\displaystyle g(x)=0$ then there exists some $\displaystyle \xi\in(x,x_0)\subseteq(x_0-\delta,x_0+\delta)$ such that $\displaystyle g'(\xi)=0$ but this contradicts that $\displaystyle g$ is never zero on $\displaystyle (x_0-\delta,x_0+\delta)$ and similarly $\displaystyle g(x)\ne 0$ for any $\displaystyle x>x_0\in(x_0-\delta,x_0+\delta)$. Make sense?
• Apr 10th 2011, 12:53 PM
paulrb
Oh I get it now, thank you very much!