Little $o$ and derivative

Question

I know that when $f(x)$ is differentiable,

$$f(x+c)=f(x)+f'(x)c+o(c).$$

I'm trying to express the following $$(h(a+c)-h(a))+(g(b-c)-g(b))$$ with little $o$, my attempt is as follows

$$h'(a)c+o(c)-g'(b)c+o(c)=c(h'(a)-g'(b)+2o(1)).$$

Is this correct?

Answer 1

Yes and simply we can write for $c\to 0$

$$h'(a)c+o(c)-g'(b)c+o(c)=c(h'(a)-g'(b)+\color{red}{o(1)})$$

since $$2\cdot o(1)=o(1)$$