Answer
$\lim\limits_{\Delta x\to 0}\dfrac{(x+\Delta x)^3-x^3}{\Delta x}=3x^2.$
Work Step by Step
$\lim\limits_{\Delta x\to 0}\dfrac{(x+\Delta x)^3-x^3}{\Delta x}=\dfrac{x^3+3x^2(\Delta x)+3x(\Delta x)^2+(\Delta x)^3-x^3}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\dfrac{\Delta x(3x^2+3x(\Delta x)+(\Delta x)^2)}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}3x^2+3x(\Delta x)+(\Delta x)^2=3x^2+3x(0)+(0)^2=3x^2.$
