Proving a point is a local minimum I have a rather basic question. I have a function $f:R \rightarrow R$, and I want to show a point, $x^*$, is local minimum, i.e., $f(x^*+\delta)  \geq f(x^*), \ \delta \to 0$.
I can show that: $f(x^* + \delta) = f(x^*) + g(\delta) + h(\delta^2)$, where $g$ and $h$ are some real-valued functions. I also can show that $g(\delta)  \geq 0$. 
Now, my question is: can I conclude that as $\delta \to 0$ I have $f(x^*+\delta)  \geq f(x^*)$ and hence, $x^*$ is a local minimum?
My current understanding is that I can, because $\frac{h(\delta^2)}{g(\delta)} \to 0$ as $\delta \to 0$, and thus I can ignore $h$. Is that right?

Consider $f(x)=x^2\sin \frac{1}{x^2}, g(x)=0,h(x)=x\sin \frac{1}{x}.$ At $x=0$ define $f(0)=h(0)=0.$ So, at $x^*=0$ you have
$$f(x^* + \delta) = f(x^*) + g(\delta) + h(\delta^2).$$
However, $f$ has not a local minimum at $x^*=0$ since it changes its sign in any neigbourhood containing $x^*=0.$
Note that $f$ and $h$ are continuous but not derivable at $x=0.$ If we modify slightly the functions, $f(x)=x^4\sin \frac{1}{x^2}, g(x)=0,h(x)=x^2\sin \frac{1}{x}$ with $f(0)=h(0)=0,$ they are also derivable at $x=0.$
Even we can have $g(x)>0.$ Consider $f(x)=x^2\sin \frac{1}{x^2},$ $g(x)=e^{-x^2},$ $h(x)=x\sin \frac{1}{x}-e^{-x}.$ At $x=0$ define $f(0)=h(0)=0.$
