Question

помогите плз на паре сижу
lim x->9 (x-9)/(√x-3)

lim x->3 (4x² -11x-3)/(5x² -16x+3)

lim x->бесконечности
(20 x² -5x+4)/(20x-5)

Answer 1

1)
$lim_{xto9}frac{x-9}{sqrt{x}-3}=lim_{xto9}frac{(x-9)(sqrt{x}+3)}{(sqrt{x}-3)(sqrt{x}+3)}=lim_{xto9}frac{(x-9)*(sqrt{x}+3)}{x-9}=\=lim_{xto9}(sqrt{x}+3)=3+3=6$


2)
$lim_{xto3} frac{(4x^2 -11x-3)}{(5x^2 -16x+3)}=lim_{xto3}frac{4(x+frac{1}{4})(x-3)}{5(x-frac{1}{5})(x-3)}=lim_{xto3}frac{4x+1}{5x-1}=frac{13}{14}$


3)
$lim_{xtoinfty}frac{(20x^2 -5x+4)}{(20x-5)}=lim_{xtoinfty}frac{x(20x-5)}{(20x-5)}+lim_{xtoinfty}frac{4}{20x-5}=\=lim_{xtoinfty}x+0=infty$