Question

Срочно пожалуйста , буду благодарна.............

Answer 1

$displaystyle lim_{x to+ infty} frac{7x+4}{3x^3-5x+1}=frac{infty}{infty}lim_{x to+ infty} frac{x^3(frac{7}{x^2}^{to 0}+frac{4}{x^3}^{to 0})}{x^3(3-frac{5}{x^2}^{to 0}+frac{1}{x^3}^{to 0})}=lim_{x to+ infty}frac{0}{3}=0$

$displaystyle lim_{x to 1}frac{x^2-2x+1}{7x-5-2x^2}=frac{0}{0}=lim_{x to 1}frac{(x-1)^2}{(x-1)(-2x+5)}=lim_{x to 1}frac{(x-1)}{(-2x+5)}=\=0$

$displaystyle lim_{x to 0}frac{2-sqrt{x^2+4}}{3x^2}=frac{0}{0}=lim_{x to 0}frac{2-sqrt{x^2+4}}{3x^2}*frac{2+sqrt{x^2+4}}{2+sqrt{x^2+4}}=\=lim_{x to 0}frac{-x^2}{3x^2(2+sqrt{x^2+4})}=lim_{x to 0}frac{-1}{3(2+sqrt{x^2+4})}=-frac{1}{12}$