I'll let h = ax, so the limit is
[tex]\displaystyle\lim_{h\to0}\frac{(x+h)^2-2(x+h)+1-(x^2-2x+1)}h[/tex]
i.e. the derivative of [tex]x^2-2x+1[/tex].
Expand the numerator to see several terms that get eliminated:
[tex](x+h)^2-2(x+h)+1-(x^2-2x+1)=x^2+2xh+h^2-2x-2h+1-x^2+2x-1=2xh+h^2-2h[/tex]
So we have
[tex]\displaystyle\lim_{h\to0}\frac{2xh+h^2-2h}h[/tex]
Since h ≠ 0 (because it is approaching 0 but never actually reaching 0), we can cancel the factor of h in both numerator and denominator, then plug in h = 0:
[tex]\displaystyle\lim_{h\to0}(2x+h-2)=\boxed{2x-2}[/tex]
