Proving the sum of two differentiable functions is also differentiable I have to prove that

If $f$ and $g$ are differentiable at some point $c$, i.e. $f'(c)$ and
  $g'(c)$ exist, then $(f+g)(x)$ is also differentiable, i.e. $(f+g)'(x)$
  exists.

In all the answers given everywhere, people have shown just that $(f+g)'(x)=f'(x)+g'(x)$. But that doesn't really show that the function is differentiable in a rigorous way (Does it?). I mean that shouldn't we use the  $\epsilon-\delta$ to show actually that the limit exists?
I tried to show that given
\begin{gather}
0 < \lvert x-c \rvert < \delta_1 \implies
\Biggl\lvert\frac{f(x)-f(c)}{x-c} - f'(c)\Biggr\rvert < \epsilon_1 \\
0 < \lvert x-c \rvert < \delta_2 \implies
\Biggl\lvert\frac{g(x)-g(c)}{x-c} - g'(c)\Biggr\rvert < \epsilon_2
\end{gather}
we somehow need to show that if we set
$$ \delta = \min\{\delta_1, \delta_2\} $$
then
$$ \forall \lvert x - \delta \rvert < 0, \qquad
\Biggl\lvert\frac{(f+g)(x)-(f+g)(c)}{x-c}-L\Biggr\rvert < \epsilon$$
But I can't proceed to prove this.
 A: $0<|x-c|<\delta$ 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.
A: 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.
A: We can also use that since $f$ and $g$ are differentiable at $x=c$ that is


*

*$f(c+h)=f(c)+h\cdot f'(c)+o(h)$

*$g(c+h)=g(c)+h\cdot g'(c)+o(h)$
then we have
$$ \begin{split}
(f+g)(c+h)
&= f(c+h)+g(c+h) \\
&= f(c)+g(c)+h\cdot(f'(c)+ g'(c))+o(h) \\
&= (f+g)(c)+h\cdot (f+g)'(c)+o(h) \end{split} $$
that is also $(f+g)$ is differentiable at $x=c$.
A: 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}
