Question

найти предел. 
$\lim_{x \to \infty} x^{\frac{1}{3} }((x+1)^{\frac{2}{3} }-(x-1)^{\frac{2}{3} })$

Answer 1

Умножим числитель и знаменатель на $((x+1)^{\frac{4}{3}}+(x+1)^{\frac{2}{3}}(x-1)^{\frac{2}{3}}+(x-1)^{\frac{4}{3}})$ чтобы получить разность кубов

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

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