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.
Best Answer
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}