Answer
$f'(x) = 2x-3$
Work Step by Step
$f(x) = x^{2} - 3x +4$
$f'(x) = \lim\limits_{z \to x}\frac{(z^{2} - 3z +4) - x^{2} + 3x -4}{z-x}$
$f'(x) = \lim\limits_{z \to x}\frac{z^{2} - 3z - x^{2} + 3x}{z-x}$
$f'(x) = \lim\limits_{z \to x}\frac{(z+x)(z-x)-3(z-x)}{z-x}$
$f'(x) = \lim\limits_{z \to x}(z+x-3)$
$f'(x) = 2x-3$