Proving the sum of two differentiable functions is also differentiable

It's easier to show that if you notice that $L = f'(c) + g'(c)$, then \begin{eqnarray} \left|\frac{(f+g)(x) - (f+g)(c)}{x - c} - (f'(c) + g'(c))\right| &\leq& \left| \frac{f(x) - f(c)}{x - c} - f'(c) \right| + \left| \frac{g(x) - g(c)}{x - c} - g'(c) \right| \\&<& \epsilon_{1} + \epsilon_{2}. \end{eqnarray}

$0<|x-c|<\delta$ implies $$|\frac{f(x)+g(x)-f(c)-g(c)}{x-c}-(f'(c)+g'(c))|$$ $$\le |\frac{f(x)-f(c)}{x-c}-f'(c)|+|\frac{g(x)-g(c)}{x-c}-g'(c)|$$ by the triangle inequality.

If $f(x)$ is differentiable at $c$ then by definition, $f'(c)=\lim\limits_{h\to 0}\dfrac{f(c+h)-f(c)}{h}.$ Similarly, since $g(x)$ is differentiable at $c,$ then $g'(c) = \lim\limits_{h\to 0}\dfrac{g(c+h)-g(c)}{h}.$ By definition, $(f + g)(x)=f(x)+g(x).$ Now we want to show that $$\lim\limits_{h\to 0}\dfrac{[f(c+h)+g(c+h)]-[f(c)+g(c)]}{h}=f'(c)+g'(c).$$ Simply manipulate the terms and use limit properties to get

$$\lim\limits_{h\to 0}\dfrac{[f(c+h)+g(c+h)]-[f(c)+g(c)]}{h}=\lim\limits_{h\to 0}\dfrac{f(c+h)-f(c)}{h} -\lim\limits_{h\to 0}\dfrac{g(c+h)-g(c)}{h}$$ $$=f'(c)+g'(c)$$ by the above definition, as required.