Помогите решить, пожалуйста

1)
$\lim_{x \to \infty} \frac{ \frac{7x}{x}+ \frac{9}{x} }{ \frac{6x}{x}- \frac{1}{x} }= \lim_{x \to \infty} \frac{7+ \frac{9}{x} }{6- \frac{1}{x} }= \frac{7}{6}=1 \frac{1}{6}$

2)
$\lim_{x \to \infty} \frac{- \frac{2x}{x^2}- \frac{1}{x^2} }{ \frac{3x^2}{x^2}- \frac{4x}{x^2}+ \frac{1}{x^2} }= \lim_{x \to \infty} \frac{- \frac{2}{x}- \frac{1}{x^2} }{3- \frac{4}{x}+ \frac{1}{x^2} }= \frac{0}{3}=0$

3)
$\lim_{x \to \infty} \frac{ \frac{10x^2}{x^2}+ \frac{4x}{x^2}- \frac{3}{x^2} }{ \frac{5x^2}{x^2}+ \frac{2x}{x^2}+ \frac{1}{x^2} }= \lim_{x \to \infty} \frac{10+ \frac{4}{x}- \frac{3}{x^2} }{5+ \frac{2}{x}+ \frac{1}{x^2} }= \frac{10}{5}=2$

4)
$\frac{7*(- \frac{1}{3} )-14}{21*(- \frac{1}{3} )+2}= \frac{- \frac{7}{3} - \frac{42}{3} }{-7+2}=- \frac{49}{3}* \frac{1}{5}=- \frac{49}{15}=-3 \frac{4}{15}$