Question

Умоляю, помогите!!!!!!

Answer 1

$lim_{xto -infty}frac{1+sqrt{2x^2-1}}{x}=lim_{xto -infty}frac{frac{1}{x}+sqrt{2-frac{1}{x^2}}}{1}=sqrt2\\lim_{xto 0}frac{sin3x}{3-sqrt{2x+9}}=limfrac{3x(3+sqrt{2x+9})}{9-(2x+9)}=lim_{xto 0}frac{3(3+sqrt{2x+9})}{-2}=-frac{18}{2}=-9\\lim_{xto 1}frac{1-x^2}{1-sqrt{x}}=lim_{xto 1}frac{(1-x)(1+x)}{1-sqrt{x}}=lim_{xto 1}frac{(1-sqrt{x})(1+sqrt{x})(1+x)}{1-sqrt{x}}=\\=lim_{xto 1}(1+sqrt{x})(1+x)=2cdot 2=4$

$lim_{xto -1}frac{sin(x+1)}{1-x^2}=lim_{xto -1}frac{x+1}{(1-x)(1+x)}=lim_{xto -1}frac{1}{1-x}=frac{1}{2}$