Suppose we are given the following system: \begin{equation} \dot{x} = Ax(t) + g(x(t)) \end{equation} We are given that the property $\lim_{x \to 0} \frac{\|g(x)\|}{\|x\|} = 0$. Now, I want to show the origin is an equilibrium point for this system.
My attempt: Showing this is equivalent to showing $g(x) = 0$. If we use Taylor expansion: \begin{equation} g(x) = g(0) + g'(0)x + \frac{g''(0)}{2!}x^{2} + ... \end{equation} Then: \begin{equation} \frac{g(x)}{||x||} = \frac{g(0)}{||x||} + \frac{g'(0)}{||x||}x + \frac{g''(0)}{2!{||x||}}x^{2} + ... \end{equation}
\begin{equation} \frac{||g(x)||}{||x||} \leq \frac{||g(0)||}{||x||} + \frac{||g'(0)||}{||x||}||x|| + \frac{||g''(0)||}{2!{||x||}}||x||^{2} + ... \end{equation}
\begin{equation} \lim_{x \to 0}\frac{||g(x)||}{||x||} \leq \lim_{x \to 0} \{ \frac{||g(0)||}{||x||} + \frac{||g'(0)||}{||x||}||x|| + \frac{||g''(0)||}{2!{||x||}}||x||^{2} + ...\} \end{equation}
which leads to
\begin{equation} 0 \leq||g'(0)|| \end{equation}
which is not useful. Any help would be much appreciated.
I am not sure that the following is what you want.
If you assume $\displaystyle \lim_{\left\|x \right\| \to 0}\dfrac{\left\|g(x) \right\|}{\left\|x \right\|}=0$ then
for every $\epsilon>0$ there exist a $\delta>0$ such that for $\left\|x \right\|<\delta$ we have:
$\dfrac{\left\|g(x) \right\|}{\left\|x \right\|}<\epsilon$. Taking $\epsilon=1$ we obtain $\left\|g(x) \right\|<\left\|x \right\|$ for $\left\|x \right\|<\delta$.
Thus, setting $x=0$ we obtain $g(x)=0$.