Proof that if f is differentiable and g is differentiable, then f + g is differentiable (using linear approximation)

Question

I am having some difficulty understanding a proof that $(f+g)$ is differentiable. We take $f$ and $g$ as differentiable and use a definition of differentiability:

$f(x+v) - f(x) = L_x + \alpha(v)$ with $\lim\limits_{v \to 0} \frac{\alpha(v)}{||v||} = 0$

$g(x+v) - g(x) = M_x + \beta(v) $ with $\lim\limits_{v \to 0} \frac{\beta(v)}{||v||} = 0$

Then we have:

$(f+g)(x + v) - (f+g)(x) = f(x+v)+g(x+v)-f(x)-g(x)$

From the definitions:

$= L_x(v) + \alpha(v) + M_x(v) + \beta(v)$

$= (L_x + M_x)(v) + (\alpha + \beta)(v)$ (1)

We are also told that:

$\lim\limits_{v \to 0} \frac{(\alpha + \beta)(v)}{||v||} = \lim\limits_{v \to 0} \frac{\alpha(v) + \beta(v)}{||v||}$ (2)

I am not sure how to arrive at (1) and (2). I thought it might relate to $L_x$ and $M_x$ being linear transformations, but I am not sure if this is a property. Would someone be able to help me out?

Thank you for your time.

Answer 1

Both 1) and 2) are just the definitions of the sum of two functions: $(\alpha +\beta) (v)=\alpha (v)+\beta (v)$. Since sum of two linear transformations is linear 1) and 2) complete the proof of differentiability.

Answer 2

By your step we have defined

$$(\alpha + \beta)(v):=\alpha(v)+ \beta(v)$$

then it follows that

$$\lim\limits_{v \to 0} \frac{(\alpha + \beta)(v)}{||v||} = \lim\limits_{v \to 0} \frac{\alpha(v) + \beta(v)}{||v||}=\lim\limits_{v \to 0} \frac{\alpha(v)}{||v||}+\lim\limits_{v \to 0} \frac{\beta(v)}{||v||}=0+0=0$$