A limit problem I created

[lookagain]

$\displaystyle Evaluate:$


$\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)$

 

[Deleted member 4993]

$\displaystyle Evaluate:$


$\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)$

$\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)$

$\displaystyle = \ \lim_{x \to \infty}\bigg(x\sqrt[6]{1 + \frac{1}{x} } \ + \ x\sqrt[3]{1 + \frac{1}{x}} \ + \ x\sqrt{1 + \frac{1}{x}} \ - \ 3x\bigg)$

$\displaystyle = \ \lim_{x \to \infty}\bigg(x(1 + \frac{1}{6x}....) \ + \ x(1 + \frac{1}{3x}...) \ + \ x(1 + \frac{1}{2x}...) \ - \ 3x\bigg)$

$\displaystyle = \ \lim_{x \to \infty}\bigg( \frac{1}{6}.... \ + \ \frac{1}{3}... \ + \ \frac{1}{2}... \ \bigg) \ = \ 1$