Решите умоляю пределы... lim 4x^2-7x+3/3x^2-2x-1 x->1

$4x^2-7x+3=0, \\ D=1, \\ x_1=\frac{3}{4}, x_2=1, \\ 4x^2-7x+3=4(x-\frac{3}{4})(x-1)=(4x-3)(x-1), \\ 3x^2-2x-1=0, \\ D_1=4, \\ x_1=-\frac{1}{3}, x_2=1, \\ 3x^2-2x-1=3(x+\frac{1}{3})(x-1)=(3x+1)(x-1), \\ lim_{x\to1}\frac{4x^2-7x+3}{3x^2-2x-1}=lim_{x\to1}\frac{(4x-3)(x-1)}{(3x+1)(x-1)}=lim_{x\to1}\frac{(4x-3)}{(3x+1)}= \\ =\frac{(4\cdot1-3)}{(3\cdot1+1)}=\frac{1}{4}$