# Thread: Lemma that is used for L'hospital's rule

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

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

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

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

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 $x_0$ such that $g(x_0)=0$ yet $g'(x_0)\ne 0$ can you find some neighborhood of $x_0$ such that $g'$ and $g$ are neither zero anywhere on that neighborhood except $g$ is of course zero at $x_0$. Anyways..

Since $g'(x_0)\ne 0$ and $g'$ is continuous there exists an open ball $B_\delta(x_0)$ such that $g'$ does not vanish on that open ball. Now, if $g$ vanished somewhere else on that open ball besides $x_0$ you could apply Rolle's theorem to find a zero of $g'$ on the open ball contradictory to construction.

3. 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 $g'(x_0) \neq 0$, there has to be some point near $x_0$where g is not 0. Since g is continuous, all of the points between $x_0$ 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.

4. 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 $g'(x_0) \neq 0$, there has to be some point near $x_0$where g is not 0. Since g is continuous, all of the points between $x_0$ 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 $g'(x_0)\ne 0$, right? Let's assume just for ease that $g'(x_0)>0$. Then, by continuity of $g'$ there exists some $\delta>0$ such that $\displaystyle |x-x_0|<\delta\implies \left|g(x_0)-g(x)\right|<\frac{g(x_0)}{2}$ which in particular implies that if $|x-x_0|<\delta$ then $g'(x)\ne 0$ (check this). Thus, we have on the interval $(x_0-\delta,x_0+\delta)$ that $g'$ is not zero. Now, I claim that except for $x_0$ $g(x)\ne 0$ for $x\in(x_0-\delta,x_0+\delta)$. Why? If $x such that $g(x)=0$ then there exists some $\xi\in(x,x_0)\subseteq(x_0-\delta,x_0+\delta)$ such that $g'(\xi)=0$ but this contradicts that $g$ is never zero on $(x_0-\delta,x_0+\delta)$ and similarly $g(x)\ne 0$ for any $x>x_0\in(x_0-\delta,x_0+\delta)$. Make sense?

5. Oh I get it now, thank you very much!