Using this definition:
If $f$ is a function and has derivative $f'(c)$ at the point $c$ in the domain of $f$ means that if ($a_n$)$_{n=1}^{\infty}$ is any sequence converging to $c$ such that $a_n$ $\not= c$is in the domain of $f$ for all $n \in \mathbb{N},$ then: $$\left[ \frac{f(x_n)-f(c)}{x_n-c}\right]_{n=1}^{\infty}$$converges to $f'(c)$
Assuming $f$ and $g$ are differentiable functions on (a,b) with $h(x)=f(x)+g(x),$ prove that $h'(x)=f'(x)+g'(x).$
Attempt so far:
Using the definition above, $\left[ \frac{h(x_n)-h(c)}{x_n-c}\right]_{n=1}^{\infty}$ converges to $h'(c)$ <=> $\left[ \frac{f(x_n)-f(c)}{x_n-c}\right]_{n=1}^{\infty}$ + $\left[ \frac{g(x_n)-g(c)}{x_n-c}\right]_{n=1}^{\infty}$ since $h(x)=f(x)+g(x).$
I'm not sure if using the fact that If ($a_n$) converges to $L$ and ($b_n$) converges to $K$ then ($a_n+b_n$) converges to $L+K$ would help at all? I feel like this proof should be somewhat simple looking at how it is proven using the normal derivative definition, but I am having trouble coming up with an actual proof.
