Answer
$\lim\limits_{\Delta x\to0}\dfrac{3(x+\Delta x)-2-(3x-2)}{\Delta x}=3.$
Work Step by Step
$\lim\limits_{\Delta x\to0}\dfrac{3(x+\Delta x)-2-(3x-2)}{\Delta x}=\lim\limits_{\Delta x\to0}\dfrac{3x+3\Delta x-2-3x+2}{\Delta x}$
$=\lim\limits_{\Delta x\to0}\dfrac{3\Delta x}{\Delta x}=\lim\limits_{\Delta x\to0}3=3.$