I'm trying to prove the sum rule of differentiation, but stuck on a particular step. Given a function $f(x) = u(x) + v(x)$, we try to differentiate $f$ using the definition of derivatives:
$$ \begin{align} \frac{df}{dx} &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\\ &= \lim_{h \rightarrow 0} \frac{u(x + h) + v(x + h) - u(x) - v(x)}{h}\\ &= \lim_{h \rightarrow 0} \frac{u(x+h) -u(x) + v(x+h) - v(x)}{h}\\ &= \lim_{h \rightarrow 0} \Big(\frac{u(x+h) -u(x)}{h} + \frac{v(x+h) - v(x)}{h}\Big)\\ &= \lim_{h \rightarrow 0} \frac{u(x+h) -u(x)}{h} + \lim_{h \rightarrow 0} \frac{v(x+h) - v(x)}{h}\\ \end{align} $$
This is where I'm stuck. I'm not sure how to justify this last move, the distribution of the limit operator over the sum.
References
The sum rule for differentiation assumes first that both $u'(x)$ and $v'(x)$ exist, so the limits exist \begin{align*} \lim_{h\rightarrow 0}\dfrac{v(x+h)-v(x)}{h}\\ \lim_{h\rightarrow 0}\dfrac{u(x+h)-u(x)}{h}, \end{align*} now turns the basic rule for limits allows us to deduce the existence of \begin{align*} \lim_{h\rightarrow 0}\left(\dfrac{v(x+h)-v(x)}{h}+\dfrac{u(x+h)-u(x)}{h}\right) \end{align*} which the value is \begin{align*} \lim_{h\rightarrow 0}\dfrac{v(x+h)-v(x)}{h}+\lim_{h\rightarrow 0}\dfrac{u(x+h)-u(x)}{h}. \end{align*}