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I am stuck with proving the following statement. In fact, I am proving another assumption, and the proof of this would help me to proceed.

Assume that $f_1$ and $f_2$ are differentiable on the interval $(a,b)$. Then define $f=f_1+f_2$. Show that $f$ is differentiable on any point in that interval. So I guess I need to prove that the sum of two differentiable functions on an interval is differentiable

My attempt: I wanted to prove it from definition, writing

$$\lim_{h\to0}\frac{f_1(a+h)-f_1(a)}{h}+\lim_{h\to 0}\frac{f_2(a+h)-f_2(a)}{h}$$

Then I can write them under the same limit, but then I cannot proceed. Would be thankful if anyone helped me with it.

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    $\begingroup$ Write down what the definition of $f'(c)$ would be for $c \in (a,b)$. (Note that since $c \in (a,b)$, $c$ can't be $a$.) $\endgroup$ – Mike Nov 18 '14 at 23:24
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$$ \frac{f(x+h)-f(x)}{h}=\frac{f_1(x+h)+f_2(x+h)-f_1(x)-f_2(x)}{h}=\frac{f_1(x+h)-f_1(x)}{h}+\frac{f_2(x+h)-f_2(x)}{h}. $$ You are probably familiar with a standard theorem about limits that says that if $L:=\lim _{x\to a}f(x)$ exists and $M:=\lim _{x\to a}g(x)$ exists, then $\lim _{x\to a}[f+g](x)$ exists and is equal to $L+M$. Applying that result to the above equation tells us that the derivative of $f$ at $x$ exists and is equal to $f_1'(x)+f_2'(x)$.

If you need to prove the statement about limits, you'll have to do an $\epsilon$-$\delta$ argument.

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$$0 \leq \lim_{h \to 0} \dfrac{|f_1(a+h) + f_2(a+h) - f_1(a) - f_2(a) - f_1'(a) - f_2'(a)|}{h}$$ $$ \leq \lim_{h \to 0} \dfrac{|f_1(a+h) - f_1(a) - f_1'(a)|}{h} + \lim_{h \to 0}\dfrac{|f_2(a+h) - f_2(a) - f_2'(a)|}{h} = 0$$

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\begin{align} & \lim_{h\to0}\frac{f(x+h)-f(x)}h = \lim_{h\to0} \left( \frac{f_1(x+h)-f_1(x)}h + \frac{f_2(x+h)-f_2(x)} h \right) \\[8pt] = {} & \left( \lim_{h\to0} \frac{f_1(x+h)-f_1(x)} h \right) + \left( \lim_{h\to0} \frac{f_2(x+h)-f_2(x)} h \right) \end{align} This last equality is true if both of the limits on the last line exist, and it is given that they do. Saying the last equality is true means the limits involved exist. The relevant theorem is that if two limits as $h\to\text{something}$ both exist, then the limit as $h\to\text{something}$ of their sum also exists and is equal to the sum of those two limits.

The first equality above is just algebra: adding fractions.

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