Question

Помогите пожалуйста!! Срочно!

Answer 1

$lim_{ alpha to 0}frac{sin alpha }{ alpha }=1\\lim_{xto 0}frac{sin^2frac{x}{3}}{x^2}=lim_{xto 0}(frac{sinfrac{x}{3}}{frac{x}{3}}cdot frac{sinfrac{x}{3}}{frac{x}{3}}cdot 9)=1cdot 1cdot 9=9\\\lim_{ alpha to infty}(1+frac{1}{alpha })^{ alpha }=e; ; to \\lim_{xto infty}(1+frac{2}{x})^{x}=lim_{xto infty}((1+frac{2}{x})^{frac{x}{2}})^{frac{2}{x}cdot x}=e^{limfrac{2}{x}cdot x}=e^2\\\lim_{xto infty}(1+frac{1}{x})^{2x}=lim_{xto infty}((1+frac{1}{x})^{x})^2=e^2$