I'll work back wards on this one. So I start with:

$\displaystyle \lim_{k,h \to 0^{+}} \frac{f(x+h) - f(x-k)}{h+k}$

$\displaystyle = \lim_{k,h \to 0^{+}} [\frac{f(x+h) - f(x-k)}{h+k} + \frac{f(x)}{h+k} - \frac{f(x)}{h+k}]$

$\displaystyle = \lim_{k,h \to 0^{+}} \frac{f(x+h) - f(x) + f(x) - f(x-k)}{h+k} $

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{1}{h+k})(\;f(x+h) - f(x) + f(x) - f(x-k)\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{h}{h+k})(\;\frac{f(x+h) - f(x)}{h} + \frac{f(x) - f(x-k)}{h}\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; \lim_{k,h \to 0^{+}}\frac{f(x+h) - f(x)}{h} + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; f'(x) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{1}{h+k}) \cdot ( \; f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}f(x) - f(x-k)\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; \frac{1}{k}f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{k}\

$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; ( \;\lim_{k,h \to 0^{+}}\frac{1}{k} \

\cdot f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + f'(x)$

$\displaystyle = \lim_{k,h \to 0^{+}} (\frac{1}{h+k}) \cdot ( \; f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + f'(x) \cdot (\lim_{k,h \to 0^{+}} \; k) \; )$

$\displaystyle = [\lim_{k,h \to 0^{+}} \frac{1}{h+k}] \cdot ( \;\;\lim_{k,h \to 0^{+}} h+ k \;\

\cdot f'(x)$

$\displaystyle = [\lim_{k,h \to 0^{+}} \frac{h+k}{h+k}] \cdot f'(x)$

$\displaystyle = [\lim_{k,h \to 0^{+}} 1] \cdot f'(x)$

$\displaystyle = (1) \cdot [f'(x)]$

$\displaystyle = f'(x)$