TITLE: Prove $(f+g)'(x) = f'(x) + g'(x)$
QUESTION [7 upvotes]: In Rudin's textbook, "Principles of Mathematical Analysis", theorem 5.3 says:

If $f$ and $g$ are defined on $[a, b]$ and are differentiable at a point $x \in [a,b]$, then
  $$(f+g)'(x) = f'(x) + g'(x)$$

Rudin said this statement is clear by theorem 4.4, but I tried to prove it by myself. Could you tell me if my way is correct?
My proof:
\begin{align}
(f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t) - (f+g)(x)}{(t-x)}\\
&= \lim _{t\to x} \frac{f(t) + g(t) - f(x) - g(x)}{(t-x)}\\
&= \lim _{t\to x} \frac{f(t)-f(x)}{t-x} + \lim _{t\to x} \frac{g(t)- g(x)}{t-x}\\
&= f'(x) + g'(x)
\end{align}

REPLY [5 votes]: It's correct.
I think I'd add a step that says this:
$$
\lim_{t\to x}\left( \frac{f(t)-t(x)}{t-x} + \frac{g(t)-g(x)}{t-x} \right)
= \lim_{t\to x} \frac{f(t)-t(x)}{t-x} + \lim_{t\to x}\frac{g(t)-g(x)}{t-x}.
$$
The reason I'd want this to be there is:

to be explicit about the fact that that is where I relied on the equality between the limit of the sum and the sum of the limits, provided the latter two limits exist;
two make it clear at what point the proof relies on the hypothesis that $f$ and $g$ are differentiable at $x$.

You could just replace your second line in your question with the left side of what I wrote above.