Answer
$\lim\limits_{\Delta x\to 0}=\dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}=2x-2.$
Work Step by Step
$\lim\limits_{\Delta x\to 0}=\dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\dfrac{x^2+2x(\Delta x) +(\Delta x)^2-2x-2\Delta x+1-x^2+2x-1}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\dfrac{\Delta x(\Delta x+2x-2)}{\Delta x}$
$=\lim\limits_{\Delta x\to 0}\Delta x+2x-2=2x-2.$
