Prove $(f+g)'(x) = f'(x) + g'(x)$

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}

• Yep, looks good. You might want to work on your formatting though. Aug 3, 2013 at 21:55
• @adriano Yeah! I just edited formatting! Aug 3, 2013 at 21:56
• Looks good. I improved the formatting a little bit. Check out the source to see how I did it. Aug 3, 2013 at 22:01
• Just wait till you graduate to Rudin's Real + Complex Analysis, or Functional Analysis. Then everything is 'clear'! Aug 4, 2013 at 3:09

$$\lim_{t\to x}\left( \frac{f(t)-t(x)}{t-x} + \frac{g(t)-g(x)}{t-x} \right) = \lim_{t\to x} \frac{f(t)-t(x)}{t-x} + \lim_{t\to x}\frac{g(t)-g(x)}{t-x}.$$
• two make it clear at what point the proof relies on the hypothesis that $f$ and $g$ are differentiable at $x$.