L lookagain Elite Member Joined Aug 22, 2010 Messages 3,249 Jun 10, 2011 #1 \(\displaystyle Evaluate:\) \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\)
\(\displaystyle Evaluate:\) \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\)
D Deleted member 4993 Guest Jun 10, 2011 #2 lookagain said: \(\displaystyle Evaluate:\) \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\) Click to expand... \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg(x\sqrt[6]{1 + \frac{1}{x} } \ + \ x\sqrt[3]{1 + \frac{1}{x}} \ + \ x\sqrt{1 + \frac{1}{x}} \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg(x(1 + \frac{1}{6x}....) \ + \ x(1 + \frac{1}{3x}...) \ + \ x(1 + \frac{1}{2x}...) \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg( \frac{1}{6}.... \ + \ \frac{1}{3}... \ + \ \frac{1}{2}... \ \bigg) \ = \ 1\)
lookagain said: \(\displaystyle Evaluate:\) \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\) Click to expand... \(\displaystyle \lim_{x \to \infty}\bigg(\sqrt[6]{x^6 + x^5} \ + \ \sqrt[3]{x^3 + x^2} \ + \ \sqrt{x^2 + x} \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg(x\sqrt[6]{1 + \frac{1}{x} } \ + \ x\sqrt[3]{1 + \frac{1}{x}} \ + \ x\sqrt{1 + \frac{1}{x}} \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg(x(1 + \frac{1}{6x}....) \ + \ x(1 + \frac{1}{3x}...) \ + \ x(1 + \frac{1}{2x}...) \ - \ 3x\bigg)\) \(\displaystyle = \ \lim_{x \to \infty}\bigg( \frac{1}{6}.... \ + \ \frac{1}{3}... \ + \ \frac{1}{2}... \ \bigg) \ = \ 1\)
