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

• Apr 10th 2011, 01:08 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 $x_0$ 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.
• Apr 10th 2011, 01:13 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 $x_0$ 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.
• Apr 10th 2011, 01:30 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 $g'(x_0) \neq 0$, 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.
• Apr 10th 2011, 01:37 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 $g'(x_0) \neq 0$, 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?
• Apr 10th 2011, 01:53 PM
paulrb
Oh I get it now, thank you very much!