Question

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

Answer 1

 

$4x^2-7x+3=0, \ D=1, \ x_1=frac{3}{4}, x_2=1, \ 4x^2-7x+3=4(x-frac{3}{4})(x-1)=(4x-3)(x-1), \ 3x^2-2x-1=0, \ D_1=4, \ x_1=-frac{1}{3}, x_2=1, \ 3x^2-2x-1=3(x+frac{1}{3})(x-1)=(3x+1)(x-1), \ lim_{xto1}frac{4x^2-7x+3}{3x^2-2x-1}=lim_{xto1}frac{(4x-3)(x-1)}{(3x+1)(x-1)}=lim_{xto1}frac{(4x-3)}{(3x+1)}= \ =frac{(4cdot1-3)}{(3cdot1+1)}=frac{1}{4}$